# NCERT Solutions for Class 11 Maths Chapter 16 Probability

NCERT Solutions for Class 11 Maths Chapter 16 Probability: In our previous classes, you have studied the classical theory of probability as a measure of uncertainty of the various phenomenon in the random experiment. In the previous chapter, you have studied statistics. In this chapter, you will learn the statistical approach of probability and the axiomatic approach of probability. In NCERT solutions for class 11 maths chapter 16 probability, you will learn to find the probability on the basis of collected data and observations which is known as the statistical approach of probability. There are 44 questions in 3 exercises in the NCERT textbook. First, try to solve it on your own, if you are not able to do so, you can take help of CBSE NCERT solutions for class 11 maths chapter 16 probability. This chapter is very important for CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT etc. Solutions of NCERT for class 11 maths chapter 16 probability is useful to study advanced topics like probability distribution, stochastic process, mathematics statics and probability(MSP). Check all NCERT solutions to learn CBSE science and maths.

## Topics of NCERT Grade 11 Maths Chapter-16 Probability

16.1 Introduction

16.2 Random Experiments

16.3 Event

16.4 Axiomatic Approach to Probability

The complete Solutions of NCERT Class 11 Mathematics Chapter 16 is provided below:

## NCERT solutions for class 11 maths chapter 16 probability-Exercise: 16.1

A coin is tossed three times.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

The required sample space is:

S = {HHH, HHT, HTH, THH, TTH, HTT, THT, TTT}

A die is thrown two times.

When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = { $\dpi{100} (x, y) : x, y$ = 1,2,3,4,5,6}

or S = {(1,1), (1,2), (1,3), ..., (1,6), (2,1), (2,2), ..., (2,6), ..., (6, 1), (6, 2), ..., (6,6)}

A coin is tossed four times.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

The required sample space is:

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}

A coin is tossed and a die is thrown.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

And,

When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

A coin is tossed and then a die is rolled only in case a head is shown on the coin.

Let H denote Heads and T denote Tails.

For each toss, there are two possible outcomes = H or T

For H, when a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = {H1, H2, H3, H4, H5, H6, T}

2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.

Let X denote the event Room X is selected, Y denote the event Room Y is selected.

B1, B2 denote the event a boy is selected and G1, G2 denote the event a girl is selected from room X.

B3 denotes the event a boy is selected and G3, G4, G5 denote the event a girl is selected from Room Y.

The required sample space is:

S = {XB1 , XB2 , XG1 , XG2 , YB3 , YG3 , YG4 , YG5 }

One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.

Let, R denote the event the red die comes out,

W denote the event the white die comes out,

B denote the event the Blue die is chosen

When a die is thrown, the possible outcomes are = {1, 2, 3 ,4 ,5 ,6}

The required sample space is:

S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?

Let B denote the event a boy is born,

G denote the event a girl is born

The required sample space with a boy or girl in the order of their births is:

S = {BB, BG, GB, GG}

What is the sample space if we are interested in the number of girls in the family?

(ii) In a family with two child, there can be only three possible cases:

no girl child, 1 girl child or 2 girl child

The required sample is:

S = {0, 1, 2}

Given, Number of red balls =1

Number of white balls = 3

Let  R denote the event that the red ball is drawn.

And W denotes the event that a white ball is drawn.

Since two balls are drawn at random in succession without replacement,

if the first ball is red, the second ball will be white. And if the first ball is white, second can be either of red and white

The required sample space is:

S = {RW, WR, WW}

Let H denote the event that Head occurs and T denote the event that Tail occurs.

For T in first toss, the possible outcomes when a die is thrown = {1, 2, 3 ,4 ,5 ,6}

The required sample space is :

S = {HH, HT, T1, T2, T3, T4, T5, T6}

Let D denote the event the bulb is defective and N denote the event the bulb is non-defective
The required sample space is:

S = {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}

Possible outcomes when a coin is tossed = {H,T}

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

When T occurs, experiment is finished. S1 = {T}

When H occurs, a die is thrown.

If the outcome is odd ({1,3,5}), S2 = {H1, H3, H5}

If the outcome is even({2,4,6}), the die is thrown again.,

S3 = {H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

The required sample space is:

S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

Given, two slips are drawn from the box, one after the other, without replacement.

Let 1, 2, 3, 4 denote the event that 1, 2, 3, 4 numbered slip is drawn respectively.

When two slips are drawn without replacement, the first event has 4 possible outcomes and the second event has 3 possible outcomes

S = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Possible outcomes when a coin is tossed = {H,T}

If the number on the die is even {2,4,6}, the coin is tossed once.

S1 = {2H, 2T, 4H, 4T, 6H, 6T}

If the number on the die is odd {1,3,5}, the coin is tossed twice.

S2 = {1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}

The required sample space is:

S = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T}

Possible outcomes when a coin is tossed = {H,T}

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Let R1 and R2 denote the event that a red ball is drawn

and B1, B2, B3 denote the event that a blue ball is drawn

If H occurs, a die is thrown.

S1 = {H1, H2, H3, H4, H5, H6}

If T occurs, a ball from a box which contains $2$  red and $3$ black balls is drawn.

S2 = {TR1 , TR2 , TB1 , TB2 , TB3}

The required sample space is:

S = {TR1 , TR2 , TB1 , TB2 , TB3 , H1, H2, H3, H4, H5, H6}

Given, a die is thrown repeatedly untill a six comes up.

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

In the experiment 6 may come up on the first throw, or the 2nd throw, or the 3rd throw and so on till 6 is obtained.

The required sample space is:

S = {6, (1,6), (2,6), (3,6), (4,6), (5,6), (1,1,6), (1,2,6), ..., (1,5,6), (2,1,6). (2,2,6), ..., (2,5,6), ..., (5,1,6), (5,2,6), ... }

NCERT solutions for class 11 maths chapter 16 probability-Exercise: 16.2

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6}

Now,

E = event that the die shows 4 = {4}

F = event that the die shows even number = {2, 4, 6}

$\dpi{100} \cap$ F = {4} $\dpi{100} \cap$ {2, 4, 6}

= {4}  $\dpi{100} \neq \phi$

Hence E and F are not mutually exclusive event.

Question:2(i) A die is thrown. Describe the following events:

A: a number less than 7

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, A : a number less than 7

As every number on a die is less than 7

A = {1, 2, 3, 4, 5, 6} = S

Question:2(ii) A die is thrown. Describe the following events:

B: a number greater than 7

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, B: a number greater than 7

As no number on the die is greater than 7

B = $\dpi{100} \phi$

Question:​​​​​​​2(iii) A die is thrown. Describe the following events:

C: a multiple of 3.

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, C : a multiple of 3

C = {3, 6}

Question:​​​​​​​2(iv) A die is thrown. Describe the following events:

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, D : a number less than 4

D = {1, 2, 3}

Question:​​​​​​​2(v) A die is thrown. Describe the following events:

E: an even multiple greater than 4

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, E : an even number greater than 4

S1 = Subset of S containing even numbers = {2,4,6}

Therefore , E = {6}

Question:​​​​​​​2(vi). A die is thrown. Describe the following events:

F: a number not less than 3

When a die is rolled, the sample space of possible outcomes:

S = {1, 2, 3, 4, 5, 6} or {x : x $\dpi{100} \in$ N, x<7}

Given, F : a number not less than 3

F = {x: x $\dpi{100} \in$ S, x $\dpi{80} \geq$ 3 } = {3, 4, 5, 6}

Question:​​​​​​​2.(vi) A die is thrown. Describe the following events:

Also find  (a)  $A\cup B$

A = {1, 2, 3, 4, 5, 6}

B= $\dpi{100} \phi$

$\dpi{100} \therefore$ A $\dpi{80} \cup$ B = {1, 2, 3, 4, 5, 6} $\dpi{80} \cup$ $\dpi{100} \phi$ = {1, 2, 3, 4, 5, 6}

Question:​​​​​​​2.(vi)  A die is thrown. Describe the following events:

Also find (b) $A\cap B$.

A = {1, 2, 3, 4, 5, 6}

B= $\dpi{100} \phi$

$\dpi{100} \therefore$ A $\dpi{80} \cap$ B = {1, 2, 3, 4, 5, 6} $\dpi{80} \cap$ $\dpi{100} \phi$ = $\dpi{100} \phi$

Question:2.(vi) A die is thrown. Describe the following events:

Also find (c)  $B\cup C$

B= $\dpi{100} \phi$

C= {3, 6}

$\dpi{100} \therefore$ B $\dpi{80} \cup$ C = $\dpi{100} \phi$ $\dpi{80} \cup$ {3, 6} = {3, 6}

Question:​​​​​​​2.(vi) A die is thrown. Describe the following events:

(d)    Also find  $E\cap F$

E = {6}

F = {3, 4, 5, 6}

$\dpi{100} \therefore$ E $\dpi{80} \cap$ F = {6} $\dpi{80} \cap$  {3, 4, 5, 6} = {6}

Question:​​​​​​​2.(vi) A die is thrown. Describe the following events:

Also find (e) $D\cap E$

D = {1, 2, 3}

E = {6}

$\dpi{100} \therefore$ D $\dpi{80} \cap$ E = {1, 2, 3} $\dpi{80} \cap$  {6} = $\dpi{100} \phi$ (As nothing is common in these sets)

Question:​​​​​​​2.(vi)  A die is thrown. Describe the following events:

Also find  (f)  $A-C$

A = {1, 2, 3, 4, 5, 6}

C = {3, 6}

$\dpi{100} \therefore$ A - C = {1, 2, 3, 4, 5, 6} - {3, 6}  = {1, 2, 4, 5}

Question:​​​​​​​2.(vi) A die is thrown. Describe the following events:

Also find (g)  $D-E$

D = {1, 2, 3}

E = {6}

$\dpi{100} \therefore$ D - E = {1, 2, 3} - {6}  = {1, 2, 3}

Question:​​​​​​​2.(vi) A die is thrown. Describe the following events:

Also find (h) $E\cap F'$

E = {6}

F = {3, 4, 5, 6}

$\dpi{100} \therefore$ F' = {3, 4, 5, 6}' = S - F = {1, 2}

$\dpi{100} \therefore$ E $\dpi{80} \cap$ F' = {6} $\dpi{80} \cap$ {1, 2} = $\dpi{100} \phi$

Question:2.(vi) A die is thrown. Describe the following events:

Also find (i)  ${F}'$

F = {3, 4, 5, 6}

$\dpi{100} \therefore$ F' = {3, 4, 5, 6}' = S - F = {1, 2}

the sum is greater than $8$

Sample space when a die is rolled:

S = {1, 2, 3, 4, 5, 6}

Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes]

E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)}

Now,

A : the sum is greater than 8

Possible sum greater than 8 are 9, 10, 11 and 12

A = [ {(a,b): (a,b) $\dpi{100} \in$ E, a+b>8 } ]= {(3,6), (4,5), (5, 4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}

$2$ occurs on either die

Sample space when a die is rolled:

S = {1, 2, 3, 4, 5, 6}

Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes]

E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)}

Now,

B: 2 occurs on either die

Hence the number 2 can come on first die or second die or on both the die simultaneously.

B = [ {(a,b): (a,b) $\dpi{100} \in$ E, a or b = 2 } ]= {(1,2), (2,2), (3, 2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)}

the sum is at least $7$ and a multiple of $3$

Sample space when a die is rolled:

S = {1, 2, 3, 4, 5, 6}

Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes]

E = [ {(x,y): x,y $\dpi{100} \in$ S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)}

Now,

C: the sum is at least 7 and a multiple of 3

The sum can only be 9 or 12.

C = [ {(a,b): (a,b) $\dpi{100} \in$ E, a+b>6 & a+b = 3k, k $\dpi{100} \in$ I} ]= {(3,6), (6,3), (5, 4), (4,5), (6,6)}

Which pairs of these events are mutually exclusive?

For two elements to be mutually exclusive, there should not be any common element amongst them.

Also, A = {(3,6), (4,5), (5, 4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}

B = {(1,2), (2,2), (3, 2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)}

C = {(3,6), (6,3), (5, 4), (4,5), (6,6)

Now,  A $\cap$ B = $\phi$      (no common element in A and B)

Hence, A and B are mutually exclusive

Again, B $\cap$ C = $\phi$     (no common element in B and C)

Hence, B and C are mutually exclusive

Again, C $\cap$ A = {(3,6), (6,3), (5, 4), (4,5), (6,6)}

Therefore,

A and B, B and C are mutually exclusive.

mutually exclusive?

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Now,

A = Event that three heads show up = {HHH}

B = Event that two heads and one tail show up = {HHT, HTH, THH}

C = Event that three tails show up = {TTT}

D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT}

(i).  For two elements X and Y to be mutually exclusive,  X $\cap$ Y = $\phi$

$\cap$ B = {HHH} $\cap$ {HHT, HTH, THH} = $\phi$ ; Hence A and B are mutually exclusive.

$\cap$ C = {HHT, HTH, THH} $\cap$ {TTT} = $\phi$ ; Hence B and C are mutually exclusive.

$\cap$ D = {TTT} $\cap$ {HHH, HHT, HTH, HTT} = $\phi$ ; Hence C and D are mutually exclusive.

$\cap$ A = {HHH, HHT, HTH, HTT} $\cap$ {HHH} = {HHH} ; Hence D and A are not mutually exclusive.

$\cap$ C = {HHH} $\cap$ {TTT} = $\phi$ ; Hence A and C are mutually exclusive.

$\cap$ D = {HHT, HTH, THH} $\cap$ {HHH, HHT, HTH, HTT} = {HHT, HTH}  ; Hence B and D are not mutually exclusive.

simple?

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Now,

A = Event that three heads show up = {HHH}

B = Event that two heads and one tail show up = {HHT, HTH, THH}

C = Event that three tails show up = {TTT}

D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT}

(ii).If an event X has only one sample point of a sample space, it is called a simple event.

A = {HHH} and C = {TTT}

Hence, A and C are simple events.

Compound?

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, THH, TTH, TTT}

Now,

A = Event that three heads show up = {HHH}

B = Event that two heads and one tail show up = {HHT, HTH, THH}

C = Event that three tails show up = {TTT}

D = Event that a head shows on the first coin = {HHH, HHT, HTH, HTT}

(iv). If an event has more than one sample point, it is called a Compound event.

B = {HHT, HTH, THH} and D = {HHH, HHT, HTH, HTT}

Hence, B and D are compound events.

Question:5(i) Three coins are tossed. Describe

Two events which are mutually exclusive.

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, THH, TTH, TTT}

(i)

A = Event that three heads show up = {HHH}

B = Event that three tails show up = {TTT}

$\cap$ B = {HHH} $\cap$ {TTT} = $\phi$ ; Hence A and B are mutually exclusive.

Question:​​​​​​​5(ii)  Three coins are tossed. Describe

Three events which are mutually exclusive and exhaustive.

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting no tails = {HHH}

B = Getting exactly one tail = {HHT, HTH, THH}

C = Getting at least two tails = {HTT, THT, TTH}

Clearly, A  $\cap$  B  = $\phi$ ; B  $\cap$  C = $\phi$ ; C  $\cap$  A  = $\phi$

Since (A and B), (B and C) and (A and C) are mutually exclusive

Therefore A, B and C are mutually exclusive.

Also,

$\cup$ B $\cup$ C = S

Hence A, B and C are exhaustive events.

Hence, A, B and C are three events which are mutually exclusive and exhaustive.

Question:​​​​​​​5(iii). Three coins are tossed. Describe

Two events, which are not mutually exclusive.

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting at least one head = {HHH, HHT, HTH, THH, TTH}

B = Getting at most one head = {TTH, TTT}

Clearly, A  $\cap$  B  = {TTH} $\neq$ $\phi$

Hence, A and B are two events which are not mutually exclusive.

Question:​​​​​​​5.(iv)  Three coins are tossed. Describe

Two events which are mutually exclusive but not exhaustive.

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting exactly one head = {HTT, THT, TTH}

B = Getting exactly one tail = {HHT, HTH, THH}

Clearly, A  $\cap$  B  = $\phi$

Hence, A and B are mutually exclusive.

Also, A  $\cup$  B  $\neq$ S

Hence, A and B are not exhaustive.

Question:​​​​​​​5.(v)  Three coins are tossed. Describe

Three events which are mutually exclusive but not exhaustive

Sample space when three coins are tossed = [Sample space when a coin is tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Let ,

A = Getting exactly one tail = {HHT, HTH, THH}

B = Getting exactly two tails = {HTT, TTH, THT}

C = Getting exactly three tails = {TTT}

Clearly, A  $\cap$  B  = $\phi$ ; B  $\cap$  C = $\phi$ ; C  $\cap$  A  = $\phi$

Since (A and B), (B and C) and (A and C) are mutually exclusive

Therefore A, B and C are mutually exclusive.

Also,

$\cup$ B $\cup$ C = {HHT, HTH, THH, HTT, TTH, THT, TTT} $\neq$ S

Hence A, B and C are not exhaustive events.

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$.

Describe the events

$A{}'$

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

(i) Therefore, A'= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

= B : getting an odd number on the first die.

Question:6.(ii)  Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$.

Describe the events

not B

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(ii) Therefore, B'= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

= A : getting an even number on the first die.

Question:​​​​​​​6.(iii)    Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$.

Describe the events

A or B

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(iii) A or B = A $\cup$ B = {(1,1), (1,2) .... (1,6), (3,1), (3,2).... (3,6), (5,1), (5,2)..... (5,6), (2,1), (2,2)..... (2,6), (4,1), (4,2)..... (4,6), (6,1), (6,2)..... (6,6)} = S

Question:6.(iv)  Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$

Describe the events

A and B

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(iii) A and B = A $\cap$ B = A $\cap$ A' = $\phi$ (From (ii))

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$

Describe the events

A but not C

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b):  2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(v) A but not C = A - C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Question:​​​​​​​6.(vi)   Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$

Describe the events

B or C

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b):  2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(vi) B or C = B $\cup$ C = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

Question:​​​​​​​6.(vii) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$

Describe the events

B and C

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b):  2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(vi) B and C = B $\cap$ C = {(1, 1), (1,2), (1,3), (1,4), (3,1), (3,2)}

Question:​​​​​​​6.(viii) Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice  $\leq 5$

Describe the events

$A\cap {B}'\cap {C}'$

Sample space when two dice are thrown:

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A: getting an even number on the first die = {(a,b): a $\dpi{80} \in$ {2,4,6} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B: getting an odd number on the first die = {(a,b): a $\dpi{80} \in$ {1,3,5} and 1 $\dpi{80} \leq$ b $\dpi{80} \leq$ 6}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C: getting the sum of the numbers on the dice $\dpi{80} \leq$ 5

The possible sum are 2,3,4,5

C = {(a,b):  2 $\dpi{80} \leq$ a + b $\dpi{80} \leq$ 5} = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(viii) A $\cap$ B' $\cap$ C' = A $\cap$ A $\cap$ C'  (from (ii))

= A $\cap$ C' = A - C = {(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Question:​​​​​​​7.(i) Refer to question 6 above, state true or false: (give reason for your answer)

A and B are mutually exclusive

Here,

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(i) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

$\cap$ B = $\phi$ , since A and B have no common element amongst them.

Hence, A and B are mutually exclusive. TRUE

A and B are mutually exclusive and exhaustive

Here,

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(ii) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

$\cap$ B = $\phi$ , since A and B have no common element amongst them.

Hence, A and B are mutually exclusive.

Also,

$\cup$ B = {(2,1), (2,2).... (2,6), (4,1), (4,2).....(4,6), (6,1), (6,2)..... (6,6), (1,1), (1,2).... (1,6), (3,1), (3,2)..... (3,6), (5,1), (5,2).... (5,6)} = S

Hence, A and B are exhaustive.

TRUE

Question:​​​​​​​7.(iii) Refer to question 6 above, state true or false: (give reason for your answer)

$A=B{}'$

Here,

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

(iii) Therefore, B' = S -B = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} = A

TRUE

Question:​​​​​​​7.(iv) Refer to question 6 above, state true or false: (give reason for your answer)

A and C are mutually exclusive

Here,

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

C =  {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(iv) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

$\cap$ C = {(2,1), (2,2), (2,3), (4,1)} ,

Hence, A and B are not mutually exclusive. FALSE

$A$ and ${B}'$ are mutually exclusive.

X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

$\cap$ B' = A $\cap$ A = A   (From (iii))

$\dpi{100} \therefore$ A $\cap$ B’ $\dpi{100} \neq \phi$

Hence A and B' not mutually exclusive. FALSE

Question:​​​​​​​7.(vi)  Refer to question 6 above, state true or false: (give reason for your answer)

${A}',{B}',C$ are mutually exclusive and exhaustive.

Here,

S = {(x,y): 1 $\dpi{80} \leq$ x,y $\dpi{80} \leq$ 6}

A = {(2, 1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}

C = {(1, 1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

(vi) X and Y are mutually exclusive if and only if X $\cap$ Y = $\phi$

$\dpi{100} \therefore$ A' $\cap$ B' = B $\cap$ A = $\phi$ (from (iii) and (i))

Hence A' and B' are mutually exclusive.

Again,

$\dpi{100} \therefore$ B' $\cap$ C = A $\cap$ C $\dpi{80} \neq$ $\phi$ (from (iv))

Hence B' and C are not mutually exclusive.

Hence, A', B' and C are not mutually exclusive and exhaustive. FALSE

CBSE NCERT solutions for class 11 maths chapter 16 probability-Exercise: 16.3

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (a) $0.1$ $0.01$ $0.05$ $0.03$ $0.01$ $0.2$ $0.6$

(a) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one.

Condition (ii): Sum of probabilities = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

Therefore, the assignment is valid

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (b) $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$

(b) Condition (i): Each of the number p( $\dpi{100} \omega_i$ ) is positive and less than one.

Condition (ii): Sum of probabilities = $\dpi{100} \frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7} = 1$

Therefore, the assignment is valid

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (c) $0.1$ $0.2$ $0.3$ $0.4$ $0.5$ $0.6$ $0.7$

(c) Since sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1

Hence, Condition (ii) is not satisfied.

Therefore, the assignment is not valid

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (d) $-0.1$ $0.2$ $0.3$ $0.4$ $-0.2$ $0.1$ $0.3$

(d) Two of the probabilities p( $\omega_1$ ) and p( $\omega_5$ ) are negative, hence condition(i) is not satisfied.

Therefore, the assignment is not valid.

 Assignment $\omega _1$ $\omega _2$ $\omega _3$ $\omega _4$ $\omega _5$ $\omega _6$ $\omega _7$ (e) $\frac{1}{14}$ $\frac{2}{14}$ $\frac{3}{14}$ $\frac{4}{14}$ $\frac{5}{14}$ $\frac{6}{14}$ $\frac{15}{14}$

(e) Each of the number p( $\dpi{100} \omega_i$) is positive but p( $\dpi{100} \omega_7$) is not less than one. Hence the condition is not satisfied.

Therefore, the assignment is not valid.

Sample space when a coin is tossed twice, S = {HH, HT, TH, TT}

[Note: A coin tossed twice is same as two coins tossed at once]

$\therefore$ Number of possible outcomes n(S) = 4

Let E be the event of getting at least one tail = {HT, TH, TT}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{3}{4}$

= 0.75

A prime number will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a prime number = {2,3,5}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{3}{6}$

= 0.5

Question:3.(ii) A die is thrown, find the probability of following events:

A number greater than or equal to $3$ will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number greater than or equal to 3 = {3,4,5,6}

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{4}{6} = \frac{2}{3}$

= 0.67

Question:​​​​​​​3.(iii)  A die is thrown, find the probability of following events:

A number less than or equal to one will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number less than or equal to one = {1}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{6}$

= 0.167

Question:3.(iv) A die is thrown, find the probability of following events:

A number more than $\small 6$  will appear

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number more than 6 will appear = $\phi$

$\therefore$ n(E) = 0

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{0}{6}$

= 0

A number less than $\small 6$ will appear.

Sample space when a die is thrown, S = {1,2,3,4,5,6}

$\therefore$ Number of possible outcomes n(S) = 6

Let E be the event of getting a number less than 6 will appear = {1,2,3,4,5}

$\therefore$ n(E) = 5

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{5}{6}$

= 0.83

(a) Number of points(events) in the sample space = Number of cards in the pack = 52

Calculate the probability that the card is an ace of spades.

Number of possible outcomes, n(S) = 52

Let E be the event that the card is an ace of spades

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{52}$

The required probability that the card is an ace of spades is $\dpi{80} \frac{1}{52}$.

Question:4(c)(i) A card is selected from a pack of 52 cards.

Calculate the probability that the card is an ace

Number of possible outcomes, n(S) = 52

Let E be the event that the card is an ace. There are 4 aces.

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{4}{52} = \frac{1}{13}$

The required probability that the card is an ace is $\dpi{80} \frac{1}{13}$.

Calculate the probability that the card is black card.

Number of possible outcomes, n(S) = 52

Let E be the event that the card is a black card. There are 26 black cards. (Diamonds and Clubs)

$\therefore$ n(E) = 26

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{26}{52} = \frac{1}{2}$

The required probability that the card is an ace is $\dpi{80} \frac{1}{2}$.

The coin and die are tossed together.

The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes

Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Number of possible outcomes, n(S) = 12

(i) Let E be the event having sum of numbers as 3 = {(1, 2)}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{12}$

The required probability of having 3 as sum of numbers is $\dpi{80} \frac{1}{12}$.

The coin and die are tossed together.

The coin can have only 1 or 6 as possible outcomes and the die can have {1,2,3,4,5,6} as poosible outcomes

Sample space, S = {(x,y): x $\dpi{80} \in$ {1,6} and y $\dpi{80} \in$ {1,2,3,4,5,6}}

= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Number of possible outcomes, n(S) = 12

(ii) Let E be the event having sum of numbers as 12 = {(6, 6)}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{12}$

The required probability of having 12 as sum of numbers is $\dpi{80} \frac{1}{12}$.

There are four men and six women on the city council

$\therefore$ n(S) = n(men) + n(women) = 4 + 6 = 10

Let E be the event of selecting a woman

$\therefore$ n(E) = 6

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{6}{10} = \frac{3}{5}$

Therefore, the required probability of selecting a woman is 0.6

Here the sample space is,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}

According to question,

1.)  4 heads = 1 + 1 + 1 + 1 = Rs. 4

2.)  3 heads and 1 tail = 1 + 1 + 1 - 1.50 = Rs. 1.50

3.)  2 heads and 2 tails = 1 + 1 - 1.50 - 1.50 = - Rs. 1 : he will lose Re. 1

4.)  1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50 = - Rs. 3.50 : he will lose Rs. 3.50

5.)  4 tails = – 1.50 – 1.50 – 1.50 – 1.50 = - Rs. 6 = he will lose Rs. 6

Now, sample space of amounts corresponding to S:

S' = {4, 1.50, 1.50, 1.50, 1.50, - 1, - 1, - 1, - 1, - 1, - 1, - 3.50, - 3.50, - 3.50, - 3.50, - 6}

$\therefore$ n(S') = 12

$\therefore$ Required Probabilities are:

$\dpi{100} P(Winning\ Rs.\ 4) = \frac{n(Winning\ Rs.\ 4)}{n(S')}$  $= \frac{1}{16}$

$\dpi{100} P(Winning\ Rs.\ 1.50) = \frac{n(Winning\ Rs.\ 1.50)}{n(S')}$  $= \frac{4}{16} = \frac{1}{4}$

$\dpi{100} P(Losing\ Re.\ 1) = \frac{n(Losing\ Re.\ 1)}{n(S')}$  $= \frac{6}{16} = \frac{3}{8}$

$\dpi{100} P(Losing\ Rs.\ 3.50) = \frac{n(Losing\ Rs.\ 3.50)}{n(S')}$  $= \frac{4}{16} = \frac{1}{4}$

$\dpi{100} P(Losing\ Rs.\ 6) = \frac{n(Losing\ Rs.\ 6)}{n(S')}$  $= \frac{1}{16}$

$\small 3$  heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting 3 heads = {HHH}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{8}$

The required probability of getting 3 heads is $\dpi{80} \frac{1}{8}$.

Question:​​​​​​​8.(ii) Three coins are tossed once. Find the probability of getting

$\small 2$  heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting 2 heads = {HHT, HTH, THH}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{3}{8}$

The required probability of getting 2 heads is $\dpi{80} \frac{3}{8}$.

Question:​​​​​​​8.(iii) Three coins are tossed once. Find the probability of getting

atleast $2$ heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting atleast 2 heads = Event of getting 2 or more heads = {HHH, HHT, HTH, THH}

$\therefore$ n(E) = 4

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{4}{8} = \frac{1}{2}$

The required probability of getting atleast 2 heads is $\dpi{80} \frac{1}{2}$.

Question:8.(iv) Three coins are tossed once. Find the probability of getting

atmost $\small 2$  heads

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting atmost 2 heads = Event of getting 2 or less heads = {HHT, HTH, THH, TTH, HTT, THT}

$\therefore$ n(E) = 6

$= \frac{6}{8} = \frac{3}{4}$$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$

The required probability of getting almost 2 heads is $\dpi{80} \frac{3}{4}$.

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting no head = Event of getting only tails = {TTT}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{8}$

The required probability of getting no head is $\dpi{80} \frac{1}{8}$.

Question:8.(vi)  Three coins are tossed once. Find the probability of getting

$\small 3$ tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting 3 tails = {TTT}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{8}$

The required probability of getting 3 tails is $\dpi{80} \frac{1}{8}$.

Question:8(vii) Three coins are tossed once. Find the probability of getting

exactly two tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting exactly 2 tails = {TTH, HTT, THT}

$\therefore$ n(E) = 3

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{3}{8}$

The required probability of getting exactly 2 tails is $\dpi{80} \frac{3}{8}$.

Question:8.(viii) Three coins are tossed once. Find the probability of getting

no tail

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting no tail = Event of getting only heads = {HHH}

$\therefore$ n(E) = 1

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{1}{8}$

The required probability of getting no tail is $\dpi{80} \frac{1}{8}$.

Question:8.(ix) Three coins are tossed once. Find the probability of getting

atmost two tails

Sample space when three coins are tossed: [Same as a coin tossed thrice!]

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8]

Let E be the event of getting atmost 2 tails = Event of getting 2 or less tails = {HHT, HTH, THH, TTH, HTT, THT}

$\therefore$ n(E) = 6

$\therefore$ $\dpi{100} P(E) = \frac{n(E)}{n(S)}$  $= \frac{6}{8} = \frac{3}{4}$

The required probability of getting atmost 2 tails is $\dpi{80} \frac{3}{4}$.

Given,

P(E) = $\small \frac{2}{11}$

We know,

P(not E) = P(E') = 1 - P(E)

= $\dpi{80} 1 - \small \frac{2}{11}$

$\dpi{80} \frac{9}{11}$

Given, ‘ASSASSINATION’

No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1

No. of letters = 13

No. of vowels = {3 A's,2 I's,O} = 6

One letter is selected:

n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13

Let E be the event of getting a vowel.

n(E) = $\dpi{100} ^{6}\textrm{C}_{1}$ = 6

$\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{6}{13}$

Given, ‘ASSASSINATION’

No. of A's = 3; No. of S's = 4; No. of I's = 2; No. of N's = 2; No. of T = 1; No. of O = 1

No. of letters = 13

No. of consonants = {4 S's,2 N's,T} = 7

One letter is selected:

n(S) = $\dpi{100} ^{13}\textrm{C}_{1}$ = 13

Let E be the event of getting a consonant.

n(E) = $\dpi{100} ^{7}\textrm{C}_{1}$ = 7

$\dpi{100} \therefore$ $\dpi{100} P(E) = \frac{7}{13}$

Total numbers of numbers in the draw = 20

Numbers to be selected = 6

$\dpi{100} \therefore$ n(S) = $\dpi{100} ^{20}\textrm{C}_{6}$

Let E be the event that six numbers match with the six numbers fixed by the lottery committee.

n(E) = 1 (Since only one prize to be won.)

$\dpi{100} \therefore$ Probability of winning =

$\dpi{100} P(E) = \frac{n(E)}{n(S)}$$\dpi{100} = \frac{1}{^{20}\textrm{C}_{6}} = \frac{6!14!}{20!}$

$\dpi{100} = \frac{6.5.4.3.2.1.14!}{20.19.18.17.16.15.14!}$

$\dpi{100} = \frac{1}{38760}$

$P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$

(i) Given, $P(A)=0.5,P(B)=0.7.P(A\cap B)=0.6$

Now P(A $\cap$ B) > P(A)

(Since $\cap$ B is a subset of A, P($\cap$ B) cannot be more than P(A))

Therefore, the given probabilities are not consistently defined.

$P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$

(ii)   Given, $P(A)=0.5,P(B)=0.4,P(A\cup B)=0.8$

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ 0.8 = 0.5 + 0.4 - P(A $\cap$ B)

$\implies$ P(A $\cap$ B) = 0.9 - 0.8 = 0.1

Therefore, P(A $\cap$ B) < P(A) and P(A $\cap$ B) < P(B) , which satisfies the condition.

Hence, the probabilities are consistently defined

Question:13 Fill in the blanks in following table:

 $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $...$ (ii) $0.35$ $...$ $0.25$ $0.6$ (iii) $0.5$ $0.35$ $...$ $0.7$

We know,

$\dpi{100} P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

(i) $\dpi{100} P(A \cup B)$ =  $\dpi{100} \frac{1}{3}+\frac{1}{5}-\frac{1}{15}$ = $\dpi{100} \frac{5+3-1}{15} = \frac{7}{15}$

(ii)  $\dpi{100} 0.6 = 0.35 + P(B) - 0.25$

$\implies$ $\dpi{100} P(B) = 0.6 - 0.1 = 0.5$

(iii)  $\dpi{100} 0.7 = 0.5 + 0.35 - P(A \cap B)$

$\implies$ $\dpi{100} P(A \cap B) = 0.85 - 0.7 = 0.15$

 $P(A)$ $P(B)$ $P(A\cap B)$ $P(A\cup B)$ (i) $\frac{1}{3}$ $\frac{1}{5}$ $\frac{1}{15}$ $\dpi{100} \boldsymbol{\frac{7}{15}}$ (ii) $0.35$ 0.5 $0.25$ $0.6$ (iii) $0.5$ $0.35$ 0.15 $0.7$

Given,  $\dpi{80} P(A)=\frac{3}{5}$   and $\dpi{80} P(B)=\frac{1}{5}$

To find : $\dpi{100} P(A or B) = P(A \cup B)$

We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B) = P(A)+ P(B)$  [Since A and B are mutually exclusive events.]

$\implies$  $P(A \cup B) = \frac{3}{5}+\frac{1}{5} = \frac{4}{5}$

Therefore,   $\dpi{100} P(A \cup B) = \frac{4}{5}$

Given, $P(E)=\frac{1}{4}$,   $P(F)=\frac{1}{2}$   and    $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find : $P(E or F) = P(E \cup F)$

We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(E \cup F) =$ $\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$
Therefore,  $P(E \cup F) =$ $\frac{5}{8}$

Given, $P(E)=\frac{1}{4}$,   $P(F)=\frac{1}{2}$   and    $P(E\hspace{1mm}and\hspace{1mm}F)=\frac{1}{8}$

To find :$P(not\ E\ and\ not\ F) = P(E' \cap F')$

We know,

$P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)$

And $P(A\cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(E \cup F) =$$\frac{1}{4}+\frac{1}{2} -\frac{1}{8} = \frac{2+4-1}{8}$

$= \frac{5}{8}$

$\implies$ $P(E' \cap F') = 1 - P(E \cup F)$

$= 1- \frac{5}{8}= \frac{3}{8}$

Therefore,  $P(E' \cap F') =$ $\frac{3}{8}$

Given, $P(not\ E\ or\ not\ F) = 0.25$

For A and B to be mutually exclusive, $P(A \cap B) = 0$

Now,  $P(not\ E\ or\ not\ F) = P(E' \cup F') = 0.25$

We know,

$P(A' \cup B') = P(A \cap B)' = 1 - P(A \cap B)$
$\\ \implies 0.25 = 1 - P(E \cap F) \\ \implies P(E \cap F) = 1 - 0.25 = 0.75 \neq 0$

Hence, E and F are not mutually exclusive.

Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16

(i) $P(not\ A) = P(A') = 1 - P(A)$

$\implies$ $P(not\ A) = 1 - 0.42 = 0.58$

Therefore, P(not A) = 0.58

Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16

(ii) $P(not\ B) = P(B') = 1 - P(B)$

$\implies$ $P(not\ B) = 1 - 0.48 = 0.52$

Therefore, P(not B) = 0.52

Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16

(iii) We know,

$P(A \cup B) = P(A)+ P(B) - P(A \cap B)$

$\implies$ $P(A \cup B) = 0.42 + 0.48 - 0.16 = 0.9 - 0.16$

= 0.74

Let M denote the event that the student is studying Mathematics and B denote the event that the student is studying Biology

And total students in the class be 100.

Given, n(M) = 40 $\implies$ P(M) = $\dpi{100} \frac{40}{100} = \frac{2}{5}$

n(B) = 30$\implies$ P(M) = $\dpi{100} \frac{30}{100} = \frac{3}{10}$

n(M $\cap$ B) = 10$\implies$ P(M) = $\dpi{100} \frac{10}{100} = \frac{1}{10}$

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(M $\cup$ B) = 0.4 + 0.3 - 0.1 = 0.6

Hence, the probability that he will be studying Mathematics or Biology is 0.6

Let A be the event that the student passes the first examination and B be the event that the students passes the second examination.

P(A $\cup$ B) is probability of passing at least one of the examination.

Therefore,

P(A $\cup$ B) = 0.95 , P(A)=0.8, P(B)=0.7

We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A  $\cap$ B)

$\implies$ P(A $\cap$ B) = 0.8 + 0.7 - 0.95 = 1.5 -0.95 = 0.55

Hence,the probability that the student will pass both the examinations is 0.55

Let A be the event that the student passes English examination and B be the event that the students pass Hindi examination.

Given,

P(A)=0.75, P(A  $\cap$ B) = 0.5, P(A' $\cap$ B') =0.1

We know,

P(A' $\cap$ B') = 1 - P(A $\cup$ B)

$\implies$ P(A $\cup$ B) = 1 - 0.1 = 0.9

Also,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\implies$ P(B) = 0.9 - 0.75 + 0.5 = 0.65

Hence,the probability of passing the Hindi examination is 0.65

The student opted for NCC or NSS.

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(i) We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)