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

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

Vedclass · Accepted Answer

$1 - (\frac{2}{5})^6$

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(A) Total number of balls in the urn $= 25$. Number of balls bearing mark $'X'$ $= 10$. Number of balls bearing mark $'Y'$ $= 15$. Let $p$ be the probability of drawing a ball with mark $'X'$ and $q$ be the probability of drawing a ball with mark $'Y'$. $p = P(	ext{mark } 'X') = \frac{10}{25} = \frac{2}{5}$. $q = P(	ext{mark } 'Y') = \frac{15}{25} = \frac{3}{5}$. Since $6$ balls are drawn with replacement,the trials are Bernoulli trials. Let $Z$ be the random variable representing the number of balls with mark $'Y'$. $Z$ follows a binomial distribution with $n = 6$ and probability of success $q = \frac{3}{5}$ (since we are looking for mark $'Y'$). The probability of getting at least one $'Y'$ mark is $P(Z \geq 1) = 1 - P(Z = 0)$. $P(Z = 0) = ^{6}C_{0} \cdot p^{6} \cdot q^{0} = 1 \cdot (\frac{2}{5})^6 \cdot 1 = (\frac{2}{5})^6$. Therefore,$P(Z \geq 1) = 1 - (\frac{2}{5})^6$.

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 at least one ball will bear the $'Y'$ mark.

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