A random variable X has the following probability distribution
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | k | 2k | 2k | 3k | $k^2$ | $2k^2$ | $7k^2+k$ |
then value of E(X) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.66
The probabilities are given at,
$P(X)=0,k,2k,2k,3k,k^2,2k^2,7k^2+k$
The sum of all probabilities must be equal to 1.
$⇒0+k+2k+2k+3k+k^2+2k^2+7k^2+k$
$⇒10k^2+9k=1$
$⇒10k^2+9k-1=0$
$⇒10k^2+10k-k-1=0$
$⇒10k(k+1)-(k+1)=0$
$⇒(k+1)(10k-1)=0$
$⇒k=\frac{1}{10}$ [always positive]
$=0.1$
$∴E(X)=∑X.P(X)$
$=(0.0)+(1×0.1)+(2×0.2)+(4×0.3)+(5×0.01)+(6×2×0.01)+[7(7×0.01+0.1)]$
$=3.66$