Answer :

(i) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)


x=0, 1, 2, ………n and q = (1-p)


As the die is thrown 6 times the total number of outcomes will be .


And we know that the favourable outcomes of getting exactly 5 successes will be, either getting 2, 4 or 6 i.e., 1/6 probability of each, total, probability, p = ,q =


The probability of success is and of failure is also .


Thus, the probability of getting exactly 5 successes will be



6C5


6C5.



(ii) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)


x=0, 1, 2, ………n and q = (1-p)


As the die is thrown 6 times the total number of outcomes will be .


And we know that the favourable outcomes of getting at least 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, probability, p = , q =


The probability of success is and of failure is also .


Thus, the probability of getting at least 5 successes will be



(6C5 + 6C6)


(6C5 + 6C6).



(iii) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)


x=0, 1, 2, ………n and q = (1-p)


As the die is thrown 6 times the total number of outcomes will be .


And we know that the favourable outcomes of getting at most 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, probability of success .


The probability of success is and of failure is also .


Thus, the probability of getting at most 5 successes will be



(6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5).


(6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5).



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