MHT CET · Maths · Probability
The p.m.f of random variate \(\mathrm{X}\) is
\(\mathrm{P}(\mathrm{X})=\frac{2 x}{\mathrm{n}(\mathrm{n}+1)},x=1,2,3, \ldots \ldots, \mathrm{n} \ 0, \text { otherwise }\)
Then \(E(X)=\)
- A \(\frac{\mathrm{n}+1}{3}\)
- B \(\frac{2 \mathrm{n}+1}{3}\)
- C \(\frac{\mathrm{n}+2}{3}\)
- D \(\frac{2 \mathrm{n}-1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 \mathrm{n}+1}{3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}(\mathrm{X}) =\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, x=1,2,3, \ldots, \mathrm{n} \)
\( 0, \text { othewise }\)
\( \therefore \mathrm{E}(\mathrm{X}) =\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}} \mathrm{p}\left(x_{\mathrm{i}}\right) \)
\( =\frac{2}{\mathrm{n}(\mathrm{n}+1)}+\frac{8}{\mathrm{n}(\mathrm{n}+1)}+\ldots+\frac{2 \mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)} \)
\( =\frac{2\left(1^2+2^2+\ldots+\mathrm{n}^2\right)}{\mathrm{n}(\mathrm{n}+1)} \)
\( =\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}(\mathrm{n}+1)} \)
\( =\frac{2 \mathrm{n}+1}{3}\)
\( 0, \text { othewise }\)
\( \therefore \mathrm{E}(\mathrm{X}) =\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}} \mathrm{p}\left(x_{\mathrm{i}}\right) \)
\( =\frac{2}{\mathrm{n}(\mathrm{n}+1)}+\frac{8}{\mathrm{n}(\mathrm{n}+1)}+\ldots+\frac{2 \mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)} \)
\( =\frac{2\left(1^2+2^2+\ldots+\mathrm{n}^2\right)}{\mathrm{n}(\mathrm{n}+1)} \)
\( =\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}(\mathrm{n}+1)} \)
\( =\frac{2 \mathrm{n}+1}{3}\)
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