If the random variable X has the following distribution:
|
X |
0 |
1 |
2 |
otherwise |
|
P(X) |
k |
2k |
3k |
0 |
Match List-I with List-II:
|
List-I |
List-II |
|
(A) k |
(I) $\frac{5}{6}$ |
|
(B) P(X < 2) |
(II) $\frac{4}{3}$ |
|
(C) E(X) |
(III) $\frac{1}{2}$ |
|
(D) P(1 ≤ X ≤ 2) |
(IV) $\frac{1}{6}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
$\sum P(X)=1$
so $k+2k+3k+0=1$
$6k=1⇒k=\frac{1}{6}$
(A) k → (IV) $\frac{1}{6}$
(B) $P(X<2)=k+2k=3k=\frac{1}{2}$ (III)
(C) $\sum(X)=0×k+2k+6k+0$
$=8k=\frac{4}{3}$ (II)
(D) $P(X∈[1,2])=2k+3k=\frac{5}{6}$ (I)