Question:
If \(X\) has a binomial distribution, \(B(n,p)\) with parameters \(n\) and \(p\) such that \(P(X=2)=P(X=3)\), then \(E(X)\), the mean of variable \(X\), is
If \(X\) has a binomial distribution, \(B(n,p)\) with parameters \(n\) and \(p\) such that \(P(X=2)=P(X=3)\), then \(E(X)\), the mean of variable \(X\), is
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Equate \({}^nC_2p^2q^{n-2}\) and \({}^nC_3p^3q^{n-3}\) and solve for \(np\).
Updated On: Nov 11, 2026
- \(2-p\)
- \(3-p\)
- \(\frac{p}{2}\)
- \(\frac{p}{3}\)
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The Correct Option is B
Solution and Explanation
Let \(q=1-p\). \(P(X=r)={}^nC_r\,p^r q^{n-r}\).
\(P(X=2)=P(X=3)\Rightarrow {}^nC_2\,p^2q^{n-2}={}^nC_3\,p^3q^{n-3}\).
Dividing by \(p^2q^{n-3}\,{}^nC_2\): \(q=p\cdot\frac{{}^nC_3}{{}^nC_2}=p\cdot\frac{n-2}{3}\).
So \(3(1-p)=p(n-2)\Rightarrow 3-3p=np-2p\Rightarrow np=3-p\).
Mean \(E(X)=np=3-p\).
Answer: (2) \(3-p\)
\(P(X=2)=P(X=3)\Rightarrow {}^nC_2\,p^2q^{n-2}={}^nC_3\,p^3q^{n-3}\).
Dividing by \(p^2q^{n-3}\,{}^nC_2\): \(q=p\cdot\frac{{}^nC_3}{{}^nC_2}=p\cdot\frac{n-2}{3}\).
So \(3(1-p)=p(n-2)\Rightarrow 3-3p=np-2p\Rightarrow np=3-p\).
Mean \(E(X)=np=3-p\).
Answer: (2) \(3-p\)
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