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(B) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.Given $n = 5$ and $np + npq = 1 \cdot 8$.S

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$0 \cdot 2$

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(B) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.Given $n = 5$ and $np + npq = 1 \cdot 8$.Substituting the values,we get $5p + 5pq = 1 \cdot 8$.$5p(1 + q) = 1 \cdot 8$.Since $q = 1 - p$,we have $5p(1 + 1 - p) = 1 \cdot 8$.$5p(2 - p) = 1 \cdot 8$.$10p - 5p^2 = 1 \cdot 8$.$5p^2 - 10p + 1 \cdot 8 = 0$.Multiplying by $5$ to simplify: $25p^2 - 50p + 9 = 0$.Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$p = \frac{50 \pm \sqrt{2500 - 4(25)(9)}}{50} = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$.$p = \frac{90}{50} = 1 \cdot 8$ (not possible as $0 \le p \le 1$) or $p = \frac{10}{50} = 0 \cdot 2$.Thus,$p = 0 \cdot 2$.

If the sum of the mean and the variance of a binomial distribution for $5$ trials is $1 \cdot 8$,then $p=$

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