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

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Aicharya · Accepted Answer

We have $n=5$ and $n p+n p q=1.8$ $5 p+5 p q=1.8 \Rightarrow 5 p(1+q)=1.8$ $	herefore 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \Rightarrow 5 \mathrm{p}(2-\mathrm{p})=1.8$ $	herefore 5 p^{2}-10 p+1.8=0$ $	herefore 5 p^{2}-p-9 p+1.8=0 \quad \Rightarrow p(5 p-1)-1.8(5 p-1)=0$ $	herefore(p-1.8)(59-1)=0 \Rightarrow p=\frac{1}{5}, 1.8$ (impossible) $	herefore \quad \mathrm{p}=0.2$

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

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