question_answer 1)

An urn contzins 25 balls of which 10 ball bear a mark ‘x’ and the remaining 15 bear a mark ‘y’. A bal 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:

Let A be the  number of balls marked with ?x?, then A has the binomial distribution with                   P(all balls with bear ?x? mark)                   (ii) Let B be the number of balls marked with ?y? then B has the binomial distribution with n = 6,             P(not more than 2 will bear ?y? mark)                                                                         (iii) P (atleast one ball will bear ?y? mark)                         (iv)