Q62 / 180 · KCET 2019 · Mathematics
A random variable ' \( X \) ' has the following probability distribution
$\begin{array}{|l|l|l|l|l|l|l|l|}
\hline \mathbf{X} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline \mathbf{P}(\mathbf{X}) & \mathrm{k}-1 & 3 \mathrm{k} & \mathrm{k} & 3 \mathrm{k} & 3 \mathrm{k}^{2} & k^{2} & k^{2}+\mathrm{k} \\
\hline
\end{array}$
Then the value of \( k \) is
$\begin{array}{|l|l|l|l|l|l|l|l|} \hline \mathbf{X} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathbf{P}(\mathbf{X}) & \mathrm{k}-1 & 3 \mathrm{k} & \mathrm{k} & 3 \mathrm{k} & 3 \mathrm{k}^{2} & k^{2} & k^{2}+\mathrm{k} \\ \hline \end{array}$
Then the value of \( k \) is
- A\( -2 \)
- B\( \frac{1}{10} \)
- Correct answerC\( \frac{1}{5} \)
- D\( \frac{2}{7} \)
Solution
(C)
$\Sigma P_{i}=1$
$K-1+3 K+k+3 K+3 K^{2}+K^{2}+K^{2}+K=1$
$5 K^{2}+9 K-2=0$
$5 K^{2}+10 K-K-2=0$
$5 K(K+2)-1(K+2)=0$
$(5 K-1)(K+2)=0$
$\mathrm{~K}=\frac{1}{5},-2(\mathrm{~K}=-2$ is not possible).
Hence, the correct answer is option: 'C'.