Q80 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

A random variable \(X\) has the probability distribution \(\begin{array}{lllllll} \hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P\left(X=x_i\right) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\ \hline \end{array}\) If \(A=\left\{x_i / x_i\right.\) is a prime number}, \(B=\left\{x_i / x_i < 4\right\}\) are two events, then \(P(A \cup B)=\)

  • A
    0.31
  • B
    0.62
  • Correct answer
    C
    0.82
  • D
    0.41

Solution

The given probability distribution for random variable \(x\) \(\begin{array}{lllllll} \hline x=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P\left(X=x_i\right) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\ \hline \end{array}\) and two events \(A=\left\{x_i \mid x_i\right.\) is a prime number \(\}\) \(=\{2,3,5\}\) and \(B=\left\{x_i \mid x_i < 4\right\}=\{1,2,3\}\) So, \(\begin{aligned} & P(A \cup B)=P(X=1)+P(X=2)+P(X=3)+P(X=5) \\ & =0.2+0.3+0.12+0.2=0.82 \end{aligned}\) Hence, option (3) is correct.

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

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