Answer :

(i) Using Bernoulli’s Trial P(Success=x) = ^{n}C_{x}.p^{x}.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

⇒ ^{6}C_{5}

⇒ ^{6}C_{5.}

⇒

(ii) Using Bernoulli’s Trial P(Success=x) = ^{n}C_{x}.p^{x}.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

⇒ (^{6}C_{5} + ^{6}C_{6})

⇒ (^{6}C_{5} + ^{6}C_{6})_{.}

⇒

(iii) Using Bernoulli’s Trial P(Success=x) = ^{n}C_{x}.p^{x}.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

⇒ (^{6}C_{0} + ^{6}C_{1} + ^{6}C_{2} + ^{6}C_{3} + ^{6}C_{4} + ^{6}C_{5}).

⇒ (^{6}C_{0} + ^{6}C_{1} + ^{6}C_{2} + ^{6}C_{3} + ^{6}C_{4} + ^{6}C_{5})_{.}

⇒

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