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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, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (iii) We know,     = 0.74
Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (ii)   Therefore, P(not B) = 0.52
Given, P(A) = 0.42,  P(B) = 0.48 and  P(A and B) = 0.16 (i)   Therefore, P(not A) = 0.58
Given,  For A and B to be mutually exclusive,  Now,  We know,  Hence, E and F are not mutually exclusive.
Given, ,      and     To find : We know, And     Therefore,
Given, ,      and     To find : We know,        Therefore,
Given,     and  To find : We know,    [Since A and B are mutually exclusive events.]   Therefore,
We know,    (i)  =   =  (ii)     (iii)       (i) (ii) 0.5 (iii) 0.15
(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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