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TS EAMCET · Maths · Probability

A random variable \(\mathrm{X}\) has the following probability distribution \begin{array}{|l|c|c|c|c|c|c|c|c|c|} \hline \mathbf{X}=\mathbf{x}_{\mathrm{i}}: & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \ \hline \mathbf{P}\left(\mathbf{X}=\mathbf{x}_{\mathrm{i}}\right): & 10 \mathrm{k} & 9 \mathrm{k} & 8 \mathrm{k} & 8 \mathrm{k} & 6 \mathrm{k} & 5 \mathrm{k} & 4 \mathrm{k} & 3 \mathrm{k} & \mathrm{k} \ \hline \end{array} where \(\mathrm{k}\) is a real number If \(\mathrm{A}=\left\{x_i / x_i\right.\) is a prime number \(\}\) and \(\mathrm{B}=\left\{x_i / x_i>5\right\}\) are two events, then \(P(A \cup B)=\)

  1. A \(\frac{2}{3}\)
  2. B \(\frac{4}{9}\)
  3. C \(\frac{1}{27}\)
  4. D \(\frac{5}{6}\)
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(A) \(\frac{2}{3}\)

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