A random variable X has the following probability distribution:
|
X |
0 |
1 |
2 |
otherwise |
|
P(X) |
k |
2k |
3k |
0 |
Then:
(A) $k=\frac{1}{6}$
(B) $P(X<2)=\frac{1}{2}$
(C) $E(X)=\frac{3}{4}$
(D) $P(1<X<2)=\frac{5}{6}$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Since sum of all probabilities must be 1.
$k+2k+3k+0=1$
$⇒6k=1$
$⇒k=\frac{1}{6}$ → (A)
$P(X<2)=P(X=0)+P(X=1)$
$=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$ → (B)