Q67 / 120 · VITEEE 2017 · 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 $\mathrm{F}=\{\mathrm{X} < 4\}$, then $\mathrm{P}(\mathrm{E} \cup \mathrm{F})$ is

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

Solution

$$ \begin{array}{l} \begin{array}{l} P(E)=P(2 \text { or } 3 \text { or } 5 \text { or } 7) \\ =0.23+0.12+0.20+0.07=0.62 \\ P(F)=P(1 \text { or } 2 \text { or } 3) \\ =0.15+0.23+0.12=0.50 \\ P(E \cap F)=P(2 \text { or } 3) \\ =0.23+0.12=0.35 \\ \therefore P(E U F)=P(E)+P(F)-P(E \cap F) \\ =0.62+0.50-0.35=0.77 \end{array} \end{array} $$

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

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