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

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(NONE) Total number of balls in the urn $= 25$. Balls bearing mark $'X'$ $= 10$. Balls bearing mark $'Y'$ $= 15$. Let $p$ be the probability of drawin

Vedclass · Accepted Answer

$7 	imes (\frac{3}{5})^4$

Vedclass · Answer

(NONE) 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 $'Y'$. $p = P(	ext{ball bearing mark } 'Y') = \frac{15}{25} = \frac{3}{5}$. Let $q$ be the probability of drawing a ball with mark $'X'$. $q = P(	ext{ball bearing mark } 'X') = \frac{10}{25} = \frac{2}{5}$. Since $6$ balls are drawn with replacement,the number of trials $n = 6$. Let $Z$ be the random variable representing the number of balls with $'Y'$ mark. $Z$ follows a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{3}{5}$. The probability is given by $P(Z = z) = ^nC_z p^z q^{n-z}$. We need to find $P(Z \leq 2) = P(Z = 0) + P(Z = 1) + P(Z = 2)$. $P(Z = 0) = ^6C_0 (\frac{3}{5})^0 (\frac{2}{5})^6 = 1 	imes 1 	imes (\frac{2}{5})^6 = \frac{64}{15625}$. $P(Z = 1) = ^6C_1 (\frac{3}{5})^1 (\frac{2}{5})^5 = 6 	imes \frac{3}{5} 	imes \frac{32}{3125} = \frac{576}{15625}$. $P(Z = 2) = ^6C_2 (\frac{3}{5})^2 (\frac{2}{5})^4 = 15 	imes \frac{9}{25} 	imes \frac{16}{625} = \frac{2160}{15625}$. $P(Z \leq 2) = \frac{64 + 576 + 2160}{15625} = \frac{2800}{15625} = \frac{112}{625}$. Note: The original options provided were incorrect. The calculated result is $\frac{112}{625}$.

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 not more than $2$ will bear $'Y'$ mark.

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