A discrete random variable X has the following probability distribution:
|
X: |
0 |
1 |
2 |
3 |
4 |
5 |
|
P(X): |
b |
3b |
5b |
3b |
4b |
6b |
The value of b is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) - $\frac{1}{22}$
$\sum P(X=x) = 1$
$b + 3b + 5b + 3b + 4b + 6b = 1$
$22b = 1$
$b = \frac{1}{22}$
The value of $b$ is $\frac{1}{22}$.