An urn contains 25 balls numbered 1 to 25. Su | vedclass

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(B) Total balls = $25$.Odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.Probability of s

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$\frac{312}{625}$

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(B) Total balls = $25$.Odd numbers between $1$ and $25$ are ${1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}$,which are $13$ in total.Probability of success $(p)$ = $\frac{13}{25}$.Probability of failure $(q)$ = $1 - p = 1 - \frac{13}{25} = \frac{12}{25}$.Since $2$ balls are drawn with replacement,this follows a binomial distribution with $n = 2$.The probability of getting exactly one success is given by $P(X = 1) = ^nC_1 \cdot p^1 \cdot q^{n-1} = 2 \cdot p \cdot q$.$P(X = 1) = 2 	imes \frac{13}{25} 	imes \frac{12}{25} = \frac{312}{625}$.

An urn contains $25$ balls numbered $1$ to $25$. Suppose an odd number is considered a 'success'. $2$ balls are drawn from the urn with replacement. Find the probability of getting exactly one success.

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