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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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then $\mathrm{p}=$
Solution:
1245 Upvotes
Verified Answer
The correct answer is:
$0 \cdot 2$
(B)
We have $n=5$ and $n p+n p q=1.8$
$5 p+5 p q=1.8 \Rightarrow 5 p(1+q)=1.8$
$\therefore 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \Rightarrow 5 \mathrm{p}(2-\mathrm{p})=1.8$
$\therefore 5 p^{2}-10 p+1.8=0$
$\therefore 5 p^{2}-p-9 p+1.8=0 \quad \Rightarrow p(5 p-1)-1.8(5 p-1)=0$
$\therefore(p-1.8)(59-1)=0 \Rightarrow p=\frac{1}{5}, 1.8$ (impossible)
$\therefore \quad \mathrm{p}=0.2$
We have $n=5$ and $n p+n p q=1.8$
$5 p+5 p q=1.8 \Rightarrow 5 p(1+q)=1.8$
$\therefore 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \Rightarrow 5 \mathrm{p}(2-\mathrm{p})=1.8$
$\therefore 5 p^{2}-10 p+1.8=0$
$\therefore 5 p^{2}-p-9 p+1.8=0 \quad \Rightarrow p(5 p-1)-1.8(5 p-1)=0$
$\therefore(p-1.8)(59-1)=0 \Rightarrow p=\frac{1}{5}, 1.8$ (impossible)
$\therefore \quad \mathrm{p}=0.2$
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