A discrete random variable X has the following probability distribution :
| X | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X) | $4c^2$ | $3c^2$ | $2c^2$ | $c^2$ | $c$ | $2c$ |
The value of $P(X≤2)$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{9}{25}$
The sum of all probabilities must be equal to 1:
$P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=1$
$⇒4c^2+3c^2+2c^2+c^2+c+2c=1$
$c=\frac{-3±7}{20}⇒c=\frac{4}{20}=0.2$ (Probabilities must be positive)
$∴P(X≤2)=4c^2+3c^2+2c^2$
$=4(0.2)^2+3(0.2)^2+2(0.2)^2$
$=0.16+0.12+0.08$
$=0.36$