A random variable X has the probability distribution : $\begin{array}{|l|l|l|l|l|l|l|l|l|} \hline x & 1 & 2…

A random variable X has the probability distribution :

$\begin{array}{|l|l|l|l|l|l|l|l|l|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\ \hline \end{array}$
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$, then $P(E \cup F)$ is
  1. 0.50
  2. 0.77
  3. 0.35
  4. 0.87

Solution

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

Asked in: MHT CET Full Test 2

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