# Q&A - Ask Doubts and Get Answers

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H Harsh Kankaria
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)

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H Harsh Kankaria
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

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H Harsh Kankaria
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

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

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H Harsh Kankaria
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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H Harsh Kankaria
Given,  No. of red marbles = 10 No. of blue marbles = 20 No. of green marbles = 30 Total number of marbles = 10 + 20 + 30 = 60 Number of ways of drawing 5 marbles from 60 marbles =  (ii). We know, The probability that at least one marble is green = 1 - Probability that no marble is green Now, Number of ways of drawing no green marbles = Number of ways of drawing only red and blue...

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H Harsh Kankaria
Given,  No. of red marbles = 10 No. of blue marbles = 20 No. of green marbles = 30 Total number of marbles = 10 + 20 + 30 = 60 Number of ways of drawing 5 marbles from 60 marbles =  (i) . Number of ways of drawing 5 blue marbles from 20 blue marbles =   Probability of drawing all blue marbles =

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H Harsh Kankaria
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

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H Harsh Kankaria
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...

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H Harsh Kankaria
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

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H Harsh Kankaria
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

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H Harsh Kankaria
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...

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H Harsh Kankaria
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

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H Harsh Kankaria
Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (iii) We know,     = 0.74

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H Harsh Kankaria
Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (ii)   Therefore, P(not B) = 0.52

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H Harsh Kankaria
Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (i)   Therefore, P(not A) = 0.58

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

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H Harsh Kankaria
Given, ,      and     To find : We know, And     Therefore,

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H Harsh Kankaria
Given, ,      and     To find : We know,        Therefore,

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H Harsh Kankaria
Given,     and  To find : We know,    [Since A and B are mutually exclusive events.]   Therefore,
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