Q29 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
If a random variable $x$ has the probability distribution
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & 0 & 2k & k & 3k & 2k^{2} & 2k & k^{2}+k & 7k^{2} \\ \hline \end{array}$
then $P(3 < x \leq 6)$ is equal to
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & 0 & 2k & k & 3k & 2k^{2} & 2k & k^{2}+k & 7k^{2} \\ \hline \end{array}$
then $P(3 < x \leq 6)$ is equal to
- A0.64
- B0.22
- Correct answerC0.33
- D0.34
Solution
Sum of all probabilities = 1
$\Rightarrow 0 + 2k + k + 3k + 2k^2 + 2k + (k^2+k) + 7k^2 = 1$
$\Rightarrow 10k^2 + 9k - 1 = 0$.
$\Rightarrow(10k-1)(k+1) = 0$
$\Rightarrow k = \frac{1}{10}$ (rejecting $k = -1$).
$P(3 < x \le 6) = P(4) + P(5) + P(6)$
$= 2k^2 + 2k + k^2 + k$
$= 3k^2 + 3k = 3(0.01) + 3(0.1) = 0.33$.
$\Rightarrow 0 + 2k + k + 3k + 2k^2 + 2k + (k^2+k) + 7k^2 = 1$
$\Rightarrow 10k^2 + 9k - 1 = 0$.
$\Rightarrow(10k-1)(k+1) = 0$
$\Rightarrow k = \frac{1}{10}$ (rejecting $k = -1$).
$P(3 < x \le 6) = P(4) + P(5) + P(6)$
$= 2k^2 + 2k + k^2 + k$
$= 3k^2 + 3k = 3(0.01) + 3(0.1) = 0.33$.
Hence, the correct answer is option: 'C'.