An urn contains 25 balls of which 10 balls bear a mark X and the remaining 15 bear a mark Y. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that(i)    all will bear X mark.(ii)   not more than 2 will bear Y mark,(iii) at least one ball will bear Y mark.(iv) the number of balls with X mark and Y mark will be equal.

Explanation

Total number of balls in the urn =25

Balls bearing mark ′X′=10

Balls bearing mark ′Y′=15

p=P(ball bearing mark′X′)=10/25

q=P(ball bearing mark′Y′)=15/25
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Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with ′Y′ mark on them in the trials.

Clearly, Z has a binomial distribution with n=6 and p=52​

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