Total numbers of numbers in the draw = 20
Numbers to be selected = 6
n(S) =
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.)
Probability of winning =

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}
n(E) = 3
The required probability of getting exactly 2 tails is .

Given, Each wheel can be labelled with 10 digits.
Number of ways of selecting 4 different digits out of the 10 digits =
These 4 digits can arranged among themselves is ways.
Number of four digit numbers without repetitions =
Number of combination that can open the suitcase = 1
Required probability of getting the right sequence to open the suitcase =

(ii)
Since 4-digit numbers greater than 5000 are to be formed,
The place digit can be filled up by either 7 or 5 in ways
Since repetition is not allowed,
The remaining 3 places can be filled by remaining 4 digits in ways.
Total number of 4-digit numbers greater than 5000 =
We know, a number is divisible by 5 if unit’s place digit is either 0 or 5.
Case 1. When digit at place is 5, the...

(i)
Since 4-digit numbers greater than 5000 are to be formed,
The place digit can be filled up by either 7 or 5 in ways
Since repetition is allowed,
Each of the remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 in ways.
Total number of 4-digit numbers greater than 5000 =
(5000 cannot be counted, hence one less)
We know, a number is divisible by 5 if unit’s place...

**8.** From the employees of a company, persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

S. No. |
Name |
Sex |
Age in years |

1. | Harish | M | 30 |

2. | Rohan | M | 33 |

3. | Sheetal | F | 46 |

4. | Alis | F | 28 |

5. | Salim | M | 41 |

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over years?

Given,
Total number of persons = 5
No. of male spokesperson = 3
No. of spokesperson who is over 35 years of age = 2
Let E be the event that the spokesperson is a male and F be the event that the spokesperson is over 35 years of age.
Since only one male is over 35 years of age,
We know,
Therefore, the probability that the spokesperson will either be a male or over 35 years of age is .

Given, P(A) = 0.54, P(B) = 0.69, P(A B) = 0.35
And, P(A B) = 0.88
(iv) P(B A') = P(B-A) = P(B) - P(A B)
= 0.69 - 0.35 = 0.34
P(B A') = 0.34

Given, P(A) = 0.54, P(B) = 0.69, P(A B) = 0.35
And, P(A B) = 0.88
(iii) P(A B') = P(A-B) = P(A) - P(A B)
= 0.54 - 0.35 = 0.19
P(A B') = 0.19

Given, P(A) = 0.54, P(B) = 0.69, P(A B) = 0.35
(i) We know, P(A B) = P(A) + P(B) - P(A B)
P(A B) = 0.54 + 0.69 - 0.35 = 0.88
P(A B) = 0.88

Given, 3 letters are put in 3 envelopes.
The number of ways of putting the 3 different letters randomly = 3!
Number of ways that at least one of the 3 letters is in the correct envelope
= No. of ways that exactly 1 letter is in correct envelope + No. of ways that 2 letters are in the correct envelope(The third is automatically placed correctly)
= No. of ways that exactly 1 letter is in...

Total number of students = 100
Let A and B be the two sections consisting of 40 and 60 students respectively.
We found out in (a) that the probability that both students are in same section is
(b) Probability that both the students are in different section =

Total number of students = 100
Let A and B be the two sections consisting of 40 and 60 students respectively.
Number of ways of selecting 2 students out of 100 students.=
(a)
If both are in section A:
Number of ways of selecting 40 students out of 100 = (The remaining 60 will automatically be in section B!)
Remaining 38 students are to be chosen out of (100-2 =) 98 students
Required...

Given, Total number of tickets sold = 10,000
Number of prizes awarded = 10
(c) If ten tickets are bought:
Number of tickets not awarded = 10000 - 10 = 9990
P(not getting a prize) =

Given, Total number of tickets sold = 10,000
Number of prizes awarded = 10
(b) If two tickets are bought:
Number of tickets not awarded = 10000 - 10 = 9990
P(not getting a prize) =

Given, Total number of tickets sold = 10,000
Number of prizes awarded = 10
(a) If one ticket is bought,
P(getting a prize) =
P(not getting a prize) = 1 - P(getting a prize)

Total number of faces of a die = 6
(iii) Number of faces with number '3' = 1
P(not 3) = 1 - P(3)
Therefore, required probability P(not 3) is

Total number of faces of a die = 6
(ii) P (1 or 3) = P (not 2) = 1 − P (2)
Therefore, required probability P(1 or 3) is 0.5

Total number of faces of a die = 6
(i) Number of faces with number '2' = 3
Therefore, required probability P(2) is 0.5

Total number of ways of drawing 4 cards from a deck of 52 cards =
We know that there are 13 diamonds and 13 spades in a deck.
Now, Number of ways of drawing 3 diamonds and 1 spade =
Probability of obtaining 3 diamonds and 1 spade

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