KCET · Maths · Probability
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
- A \( -2 \)
- B \( \frac{1}{10} \)
- C \( \frac{1}{5} \)
- D \( \frac{2}{7} \)
Answer & Solution
Correct Answer
(C) \( \frac{1}{5} \)
Step-by-step Solution
Detailed explanation
(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).
\(\Sigma P_{i}=1\)
\(K-1+3 K+k+3 K+3 K^{2}+K^{2}+K^{2}+K=1\)
\(5 K^{2}+9 K-2=0\)
\(5 K^{2}+10 K-K-2=0\)
\(5 K(K+2)-1(K+2)=0\)
\((5 K-1)(K+2)=0\)
\(\mathrm{~K}=\frac{1}{5},-2(\mathrm{~K}=-2\) is not possible).
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