An urn contains 25 balls of which 10 balls bear a mark ‘ ‘ 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 ‘ ‘ mark.
(ii) Not more than 2 will bear ‘Y’ mark.
An urn contains 25 balls of which 10 balls bear a mark ‘ ‘ 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 ‘ ‘ mark.
(ii) Not more than 2 will bear ‘Y’ mark.

Solution:
(i) Given in the question that,

Total number of balls in the urn

Number of balls bearing mark

Number of balls bearing mark

In this equation, represents the probability of balls bearing mark’ ‘, while represents the probability of balls not bearing mark’ ” ‘

and

Six balls are drawn with replacement at this point. As a result, the number of trials is represented by a Bernoulli triangle.

Assume, be the random variable that represents the number of balls bearing ‘ ‘ mark in the trials

has a binomial distribution where and

Hence, (All balls will bear mark ‘ )

(ii) Probability (Not more than 2 will bear ‘ ‘ mark)