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A random variable XX has the following probability distribution:

X01234
P(X)0k2\dfrac{k}{2}3k2\dfrac{3k}{2}k3\dfrac{k}{3}2k2k

Match the LIST-I with LIST-II

LIST-ILIST-II
A.Value of kkI.713\dfrac{7}{13}
B.Value of P(X≤3)P(X \leq 3)II.326\dfrac{3}{26}
C.Value of P(X>1)P(X > 1)III.2326\dfrac{23}{26}
D.Value of P(X<2)P(X < 2)IV.313\dfrac{3}{13}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

Sum of probabilities: 13k3=1\dfrac{13k}{3} = 1, so k=313k = \dfrac{3}{13} (IV).

P(X≤3)=7k3=713P(X\le3) = \dfrac{7k}{3} = \dfrac{7}{13} (I).

P(X>1)=23k6=2326P(X>1) = \dfrac{23k}{6} = \dfrac{23}{26} (III).

P(X<2)=k2=326P(X<2) = \dfrac{k}{2} = \dfrac{3}{26} (II).

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