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 .

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 B) = 24
Therefore, P(A) =
P(B) =
P(A B) =
(iii) Now,
Probability that the student has opted NSS but not NCC = P(B A' ) = P(B-A)
We know,
P(B-A) = P(B) - P(A B)
Hence,the probability that the student has opted NSS but not NCC is

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 B) = 24
Therefore, P(A) =
P(B) =
P(A B) =
(ii) Now,
Probability that the student has opted neither NCC nor NSS = P(A' B' )
We know,
P(A' B' ) = 1 - P(A B) [De morgan's law]
And, P(A B) = P(A)+ P(B) - P(A B)
P(A' B' )
Hence,the probability...

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 B) = 24
Therefore, P(A) =
P(B) =
P(A B) =
(i) We know,
P(A B) = P(A)+ P(B) - P(A B)
Hence,the probability that the student opted for NCC or NSS is

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 B) = 0.5, P(A' B') =0.1
We know,
P(A' B') = 1 - P(A B)
P(A B) = 1 - 0.1 = 0.9
Also,
P(A B) = P(A)+ P(B) - P(A B)
P(B) = 0.9 - 0.75 + 0.5 = 0.65
Hence,the probability of passing the Hindi examination is 0.65

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 B) is probability of passing at least one of the examination.
Therefore,
P(A B) = 0.95 , P(A)=0.8, P(B)=0.7
We know,
P(A B) = P(A)+ P(B) - P(A B)
P(A B) = 0.8 + 0.7 - 0.95 = 1.5 -0.95 = 0.55
Hence,the probability that the student will pass both the...

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 P(M) =
n(B) = 30 P(M) =
n(M B) = 10 P(M) =
We know,
P(A B) = P(A)+ P(B) - P(A B)
P(M B) = 0.4 + 0.3 - 0.1 = 0.6
Hence, the probability that he will be studying Mathematics or Biology is 0.6

Given,
For A and B to be mutually exclusive,
Now,
We know,
Hence, E and F are not mutually exclusive.

(ii) Given,
We know,
P(A B) = P(A)+ P(B) - P(A B)
0.8 = 0.5 + 0.4 - P(A B)
P(A B) = 0.9 - 0.8 = 0.1
Therefore, P(A B) < P(A) and P(A B) < P(B) , which satisfies the condition.
Hence, the probabilities are consistently defined

(i) Given,
Now P(A B) > P(A)
(Since A B is a subset of A, P(A B) cannot be more than P(A))
Therefore, the given probabilities are not consistently defined.

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) = = 13
Let E be the event of getting a consonant.
n(E) = = 7

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) = = 13
Let E be the event of getting a vowel.
n(E) = = 6

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