MHT CET · Maths · Probability
If \(X\) is a random variable with distribution given below
Then the value of \(k\) and its variance are respectively given by
- A \(\frac{1}{8}, \frac{22}{27}\)
- B \(\frac{1}{8}, \frac{23}{27}\)
- C \(\frac{1}{8}, \frac{8}{9}\)
- D \(\frac{1}{8}, \frac{3}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{8}, \frac{3}{4}\)
Step-by-step Solution
Detailed explanation
The sum of all the probabilities in a probability distribution is always unity.
\(\therefore k+3 k+3 k+k=1 \)
\( \Rightarrow 8 k=1 \)
\( \Rightarrow k=\frac{1}{8}\)
\(\mathrm{E}(\mathrm{X}) =\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \)
\( =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right)=\frac{3}{2}\)
\(\operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2\)
\(=0^2\left(\frac{1}{8}\right)+1^2\left(\frac{3}{8}\right)+2^2\left(\frac{3}{8}\right)+3^2\left(\frac{1}{8}\right)-\left(\frac{3}{2}\right)^2 \)
\( =\frac{3}{4}\)
\(\therefore k+3 k+3 k+k=1 \)
\( \Rightarrow 8 k=1 \)
\( \Rightarrow k=\frac{1}{8}\)
\(\mathrm{E}(\mathrm{X}) =\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \)
\( =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right)=\frac{3}{2}\)
\(\operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2\)
\(=0^2\left(\frac{1}{8}\right)+1^2\left(\frac{3}{8}\right)+2^2\left(\frac{3}{8}\right)+3^2\left(\frac{1}{8}\right)-\left(\frac{3}{2}\right)^2 \)
\( =\frac{3}{4}\)
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