JEE PYQ: Probability Question 31
Question 31 - 2020 (09 Jan Shift 2)
A random variable $X$ has the following probability distribution:
| $X$ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| $P(X)$ | $K^2$ | $2K$ | $K$ | $2K$ | $5K^2$ |
Then $P(X > 2)$ is equal to:
(1) $\frac{7}{12}$
(2) $\frac{1}{36}$
(3) $\frac{1}{6}$
(4) $\frac{23}{36}$
Show Answer
Answer: (4)
Solution
$\sum P(X) = 1 \Rightarrow 6K^2 + 5K - 1 = 0 \Rightarrow (6K-1)(K+1) = 0 \Rightarrow K = \frac{1}{6}$. $P(X > 2) = K + 2K + 5K^2 = \frac{1}{6} + \frac{2}{6} + \frac{5}{36} = \frac{6+12+5}{36} = \frac{23}{36}$.
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