MHT CET · Maths · Probability
If the sum of the mean and the variance of a binomial distribution for 5 trials is \(1 \cdot 8\),
then \(\mathrm{p}=\)
- A \(0 \cdot 4\)
- B \(0 \cdot 2\)
- C \(0 \cdot 8\)
- D \(0 \cdot 18\)
Answer & Solution
Correct Answer
(B) \(0 \cdot 2\)
Step-by-step Solution
Detailed explanation
(B)
We have \(n=5\) and \(n p+n p q=1.8\)
\(5 p+5 p q=1.8 \Rightarrow 5 p(1+q)=1.8\)
\(\therefore 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \Rightarrow 5 \mathrm{p}(2-\mathrm{p})=1.8\)
\(\therefore 5 p^{2}-10 p+1.8=0\)
\(\therefore 5 p^{2}-p-9 p+1.8=0 \Rightarrow p(5 p-1)-1.8\)\((5 p-1)=0\)
\(\therefore(p-1.8)(59-1)=0 \Rightarrow p=\frac{1}{5}, 1.8\) (impossible)
\(\therefore \mathrm{p}=0.2\)
We have \(n=5\) and \(n p+n p q=1.8\)
\(5 p+5 p q=1.8 \Rightarrow 5 p(1+q)=1.8\)
\(\therefore 5 \mathrm{p}[1+(1-\mathrm{p})]=1.8 \Rightarrow 5 \mathrm{p}(2-\mathrm{p})=1.8\)
\(\therefore 5 p^{2}-10 p+1.8=0\)
\(\therefore 5 p^{2}-p-9 p+1.8=0 \Rightarrow p(5 p-1)-1.8\)\((5 p-1)=0\)
\(\therefore(p-1.8)(59-1)=0 \Rightarrow p=\frac{1}{5}, 1.8\) (impossible)
\(\therefore \mathrm{p}=0.2\)
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