Question -
Answer -
We know that therepeated tosses of a dice are known as Bernoulli trials.
Let the number ofsuccesses of getting an odd number in an experiment of 6 trials be x.
Probability of gettingan odd number in a single throw of a dice (p)
Thus, q = 1 тАУ p = ┬╜
Now, here x has abinomial distribution.
Thus, P(X = x) =┬аnCx┬аqn-x┬аpx,where x = 0, 1, 2 тАжn
=┬а6Cx┬а(1/2)6-x┬а(1/2)x
=┬а6Cx┬а(1/2)6
(i) Probability ofgetting 5 successes = P(X = 5)
=┬а6C5┬а(1/2)6
= 6 ├Ч1/64
= 3/32
(ii) Probability ofgetting at least 5 successes = P(X тЙе 5)
= P(X = 5) + P(X = 6)
=┬а6C5┬а(1/2)6┬а+┬а6C5┬а(1/2)6
= 6 ├Ч1/64 + 6 ├Ч1/64
= 6/64 + 1/64
= 7/64
(iii) Probability ofgetting at most 5 successes = P(X тЙд 5)
We can also write itas: 1 тАУ P(X>5)
= 1 тАУ P(X = 6)
= 1 тАУ┬а6C6┬а(1/2)6
= 1 тАУ 1/64
= 63/64