A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0 | \( \displaystyle\frac{k}{2} \) | \( \displaystyle\frac{3k}{2} \) | \( \displaystyle\frac{k}{3} \) | 2k |
Match List-I with List-II
| List-I | List-II | ||
| (A) | Value of k | (I) | \( \displaystyle\frac{7}{13} \) |
| (B) | Value of \( P(X\le 3) \) | (II) | \( \displaystyle\frac{3}{26} \) |
| (C) | Value of \( P(X\gt 1) \) | (III) | \( \displaystyle\frac{23}{26} \) |
| (D) | Value of \( P(X\lt 2) \) | (IV) | \( \displaystyle\frac{3}{13} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
The probabilities of a distribution add to 1.
(A) Value of k
This matches (IV) \( \frac{3}{13} \).
\( \displaystyle 0+\frac{k}{2}+\frac{3k}{2}+\frac{k}{3}+2k=1 \)
\( \displaystyle \Rightarrow\ \frac{3k+9k+2k+12k}{6}=1 \)
\( \displaystyle \Rightarrow\ \frac{26k}{6}=1 \)
\( \displaystyle \Rightarrow\ k=\frac{3}{13} \)
(B) Value of \( P(X\le 3) \)
This matches (I) \( \frac{7}{13} \).
\( \displaystyle P(X\le 3)=1-P(X=4)=1-2k \)
\( \displaystyle =1-\frac{6}{13}=\frac{7}{13} \)
(C) Value of \( P(X\gt 1) \)
This matches (III) \( \frac{23}{26} \).
\( \displaystyle P(X\gt 1)=1-P(X=0)-P(X=1) \)
\( \displaystyle =1-\frac{k}{2}=1-\frac{3}{26} \)
\( \displaystyle =\frac{23}{26} \)
(D) Value of \( P(X\lt 2) \)
This matches (II) \( \frac{3}{26} \).
\( \displaystyle P(X\lt 2)=P(X=0)+P(X=1) \)
\( \displaystyle =0+\frac{k}{2}=\frac{3}{26} \)
So the matching is (A)-(IV), (B)-(I), (C)-(III), (D)-(II).