• # question_answer 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$P(E\cup F)$is A) 0.87   B)                       0.77        C)        0.35        D)        0.50

If A and B are two events associated with a random experiment. Then, $P(A\cup B)=P(A)+P(B)-P(A\cap B),$ which is equal to the atleast one of the event is occur. E = {X is a prime number} = { 2, 3, 5, 7} $P(E)=P(X=2)+P(X=3)$                                 $+P(X=5)+P(X=7)$ $\Rightarrow$$P(E)=0.23+0.12+0.20+0.07+0.62$ and $F=\{X<4\}=\{1,2,3\}$ $P(F)=P(X=1)+P(X=2)+P(X=3)$ $\Rightarrow$ $P(F)=0.15+0.23+0.12=0.5$ $E\cap F=${X is prime number as well as < 4} = {2, 3} $P(E\cap F)=P(X=2)+P(X=3)$ $=0.23+0.12=0.35$ $\therefore$Required probability, $P(E\cup F)=P(E)+P(F)-P(E\cap F)$ $P(E\cup F)=0.62+0.5-0.35$ $P(E\cup F)=0.77$