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If the sum of mean and variance of a binomial distribution for 5 trials is 1.8 , then probability of a success is
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Verified Answer
The correct answer is:
0.2
We have $n p+n p q=1.8$, where $n=5$
$$
\begin{aligned}
& \therefore \mathrm{np}(1+\mathrm{q})=1.8 \Rightarrow 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \\
& \therefore 5 \mathrm{p}(2-\mathrm{p})=1.8 \Rightarrow 10 \mathrm{p}-5 \mathrm{p}^2=1.8 \text { i.e. } \\
& 5 \mathrm{p}^2-10 \mathrm{p}+1.8=0 \Rightarrow 5 \mathrm{p}^2-9 \mathrm{p}-\mathrm{p}+1.8=0 \\
& \therefore 5 \mathrm{p}(\mathrm{p}-1.8)-1(\mathrm{p}-1.8)=0 \Rightarrow(5 \mathrm{p}-1)(\mathrm{p}-1.8)=0 \\
& \therefore \mathrm{p}=\frac{1}{5}, 1.8 \text { but } \mathrm{p} \leq 1 \Rightarrow \mathrm{p}=\frac{1}{5}=0.2
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \mathrm{np}(1+\mathrm{q})=1.8 \Rightarrow 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \\
& \therefore 5 \mathrm{p}(2-\mathrm{p})=1.8 \Rightarrow 10 \mathrm{p}-5 \mathrm{p}^2=1.8 \text { i.e. } \\
& 5 \mathrm{p}^2-10 \mathrm{p}+1.8=0 \Rightarrow 5 \mathrm{p}^2-9 \mathrm{p}-\mathrm{p}+1.8=0 \\
& \therefore 5 \mathrm{p}(\mathrm{p}-1.8)-1(\mathrm{p}-1.8)=0 \Rightarrow(5 \mathrm{p}-1)(\mathrm{p}-1.8)=0 \\
& \therefore \mathrm{p}=\frac{1}{5}, 1.8 \text { but } \mathrm{p} \leq 1 \Rightarrow \mathrm{p}=\frac{1}{5}=0.2
\end{aligned}
$$
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