JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Probability

A random variable

X

has the following probability distribution:

\begin{tabular}{ccccccc}

X

:

& 1 & 2 & 3 & 4 & 5 \\

P (X)

:

K^{2}

2 K

K

2 K

5 K^{2}

\end{tabular}

Then

P (X > 2)

is equal to:

Answer: (C)

Solution

The probabilities must sum to

1

K^{2} + 2 K + K + 2 K + 5 K^{2} = 6 K^{2} + 5 K = 1

, so

6 K^{2} + 5 K - 1 = 0

, which factors as

(6 K - 1) (K + 1) = 0

. Taking the positive root,

K = \frac{1}{6}

. Then

P (X > 2) = P (3) + P (4) + P (5) = K + 2 K + 5 K^{2} = \frac{1}{6} + \frac{2}{6} + \frac{5}{36} = \frac{6 + 12 + 5}{36} = \frac{23}{36}

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option