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 $'Y'$ $= 15$. Probability of drawing a ball with mark

Vedclass · Accepted Answer

$\left(\frac{2}{5}\right)^{6}$

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(A) Total number of balls in the urn $= 25$. Balls bearing mark $'X'$ $= 10$. Balls bearing mark $'Y'$ $= 15$. Probability of drawing a ball with mark $'X'$ is $p = \frac{10}{25} = \frac{2}{5}$. Since the ball is replaced after each draw,the trials are independent Bernoulli trials. We draw $n = 6$ balls. Let $X$ be the random variable representing the number of balls with mark $'X'$. This follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{2}{5}$. The probability that all $6$ balls bear mark $'X'$ is $P(X = 6)$. Using the binomial probability formula $P(X = k) = ^{n}C_{k} p^{k} q^{n-k}$,where $q = 1 - p = \frac{3}{5}$: $P(X = 6) = ^{6}C_{6} \left(\frac{2}{5}ight)^{6} \left(\frac{3}{5}ight)^{0} = 1 	imes \left(\frac{2}{5}ight)^{6} 	imes 1 = \left(\frac{2}{5}ight)^{6}$.

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 all will bear $'X'$ mark.

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