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Q. 5 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

(ii) not more than $2$ will bear $'Y'$ mark.

Answers (1)

best_answer

Total balls in urn = 25

Balls bearing mark 'X' =10

Balls bearing mark 'Y' =15

$P(ball\, \, bearing\, mark\, 'X')=\frac{10}{25}=\frac{2}{5}$

$P(ball\, \, bearing\, mark\, 'Y')=\frac{15}{25}=\frac{3}{5}$

$6$ balls are drawn with replacement.

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

Z has binomial distribution with n=6.

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

$P(not\, more\, than\, 2\, bear\, Y) =P(Z\leq 2)$

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

$=^6C_0 (\frac{2}{5})^{6} (\frac{3}{5})^0+^6C_1 (\frac{2}{5})^{5} (\frac{3}{5})^1+^6C_2 (\frac{2}{5})^{4} (\frac{3}{5})^2$

$= (\frac{2}{5})^{6} +6 (\frac{2}{5})^{5} (\frac{3}{5})^1+15 (\frac{2}{5})^{4} (\frac{3}{5})^2$

$= (\frac{2}{5})^{4} [(\frac{2}{5})^{2}+6 (\frac{2}{5}) (\frac{3}{5})+15(\frac{3}{5})^2]$

$= (\frac{2}{5})^{4} [\frac{4}{25}+\frac{36}{25}+\frac{135}{25}]$

$= (\frac{2}{5})^{4} [\frac{175}{25}]$

$= (\frac{2}{5})^{4} [7]$

Posted by

seema garhwal

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