If a random variable X has the following probability distribution values of X.
| X: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X): | 0 | k | 2k | 2k | 3k | $k^2$ | $2k^2$ | $7k^2+k$ |
Then, P( X ≥ 6)=
Answer & explanation
Correct answer: option 1
Since the sum of the probabilities in a probability distribution is always unity.
$∴ P(X=0)+P(X=1)+.....+P(X=7)=1$
$⇒ 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 +k =1 $
$⇒ 10k^2 + 9k - 1=0 $
$⇒ (10k -1) (k+1)=0$
$⇒ 10k - 1= 0 $ $[∵k ≥ 0 ∴ k + 1 ≠ 0]$
$⇒ k = \frac{1}{10}$
Now,
$P(X ≥ 6) =P(X= 6) + P(X = 7)$
$= 2k^2 + 7k^2 + k = 9k^2 + k =\frac{19}{100}$ $[∵k =\frac{1}{10}]$