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Question

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 note down and it is replaced. If 6 balls are drawn in this way, find the probability that

not more than 2 will bear Y mark

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Solution

It is case of Bernoulli trials with n=6. Let success be defined as drawing a ball marked X.

p=P(a success in a single draw)=1025=25

q=125=35

Clearly, Z has a binomial distribution with n=6, p=25 and q=35

P(Z=r)=nCrprqnr=6Cr(25)r(35)6r

P(not more than 2 bear mark Y)

=P(not less than 4 bear mark X)= P(atleast 4 successes)

P(4)+P(5)+P(6)=6C4p4q2+6C5p5q+6C6p6q0

=6C4(25)4(35)2+6C5(25)5(35)1+6C6(25)6(35)0=15×(25)4925+6×(25)535+1×(25)6×1=(25)4[275+3625+425]=(25)4[135+36+425]=(25)4×17525=7(25)4


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