A random variable X has the following probability distributions:
| X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
For the events E = {X is a prime number}, F = { X < 4}, the probability P(E ∪ F), is
Answer & explanation
Correct answer: option 2
We have,
$P(E)=P(X=2 \, or \, X=3 \, or \, X=5 \, or \, X = 7)$
$= P(X=2) +P(X=3) +P(X=5)+P(X=7)$
$= 0.23+0.12+0.20+0.07=0.62$
$P(F) =P(X < 4)$
$= P(X=1)+P(X=2) +P(X=3)$
$= 0.15 + 0.23 + 0.12 = 0.50$
$ P(E ∩ F)=$ P( X is a prime number less than 4)
$=P(X=2)+P(X=3) =0.23 +0.12=0.35$
$∴ P(E ∪ F)= P(E) +P(F) -P(E ∩ F)$
$= 0.62 + 0.50 - 0.35 = 0.77$