Q143 / 225 · JEE Main 2004 · Mathematics

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

  • A
    $0.87$
  • Correct answer
    B
    $0.77$
  • C
    $0.35$
  • D
    $0.50$

Solution

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

Hence, the correct answer is option: 'B'.

Relevant Questions