A random variable $X$ has the probability distribution: | $\mathrm{X}:$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |---|---|---|---|---|---|---|---|---| | $\mathrm{p}(\mathrm{X}):$ | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 | For the events $\mathrm{E}=\{\mathrm{X}$ is a prime number $\}$ and $\mathrm{F}=\{\mathrm{X}<4\}$, the probability $\mathrm{P}(\mathrm{E} \cup \mathrm{F})$ is

JEE Main · 2004 · 24 Apr 2004 · Offline · Probability

A random variable

X

has the probability distribution:

$X :$	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

}

and

F = {X < 4}

, the probability

P (E \cup F)

Answer: (B)

The prime values are

{2, 3, 5, 7}

, with probability

0.23 + 0.12 + 0.20 + 0.07 = 0.62

, and the event

X < 4

{1, 2, 3}

, with probability

0.15 + 0.23 + 0.12 = 0.50

. Their overlap

{2, 3}

has probability

0.23 + 0.12 = 0.35

. By the addition rule,

P (E \cup F) = 0.62 + 0.50 - 0.35 = 0.77

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option