Q. 54.3( 15 Votes )

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.

Answer :

(i) It is given in the question that,

Total number of balls in the urn = 25


Number of balls bearing mark ‘X’ = 10


Number of balls bearing mark ‘Y’ = 15


Let p denotes the probability of balls bearing mark ‘X’ and q denotes the probability of balls bearing mark ‘Y’



And,


Now, 6 balls are drawn with replacement. Hence, the number of trials are Bernoulli triangle.


Let us assume, Z be the random variable that represents the number of balls bearing ‘Y’ mark in the trials


Z has a binomial distribution where n = 6 and



Hence, P (All balls will bear mark ‘X’) = P (Z = 0)




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


= P (Z = 0) + P (Z = 1) + P (Z = 2)






=


(iii) Now, Probability (At least one ball will bear ‘Y’ mark) = P (Z ≥ 1)


= 1 – P (Z = 0)


=


(iv) Probability (Having equal number of balls with ‘X’ mark and ‘Y’ mark) = P (Z = 3)



=


=


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