An urn contains 25 balls of which 10 balls be | vedclass

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(A) Total number of balls in the urn $= 25$. Balls bearing mark $'X'$ $= 10$. Balls bearing mark $'Y'$ $= 15$. Let $p$ be the probability of drawing a

Vedclass · Accepted Answer

$\frac{864}{3125}$

Vedclass · Answer

(A) Total number of balls in the urn $= 25$. Balls bearing mark $'X'$ $= 10$. Balls bearing mark $'Y'$ $= 15$. Let $p$ be the probability of drawing a ball with mark $'X'$: $p = \frac{10}{25} = \frac{2}{5}$. Let $q$ be the probability of drawing a ball with mark $'Y'$: $q = \frac{15}{25} = \frac{3}{5}$. Since $6$ balls are drawn with replacement,this follows a binomial distribution with $n = 6$. We want the number of balls with $'X'$ mark and $'Y'$ mark to be equal,which means we need $3$ balls of mark $'X'$ and $3$ balls of mark $'Y'$. Using the binomial probability formula $P(Z = k) = ^{n}C_{k} p^{k} q^{n-k}$: $P(Z = 3) = ^{6}C_{3} 	imes (\frac{2}{5})^{3} 	imes (\frac{3}{5})^{3}$. $P(Z = 3) = 20 	imes \frac{8}{125} 	imes \frac{27}{125}$. $P(Z = 3) = \frac{20 	imes 216}{15625} = \frac{4320}{15625} = \frac{864}{3125}$.

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