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AP EAMCET · Maths · Probability

A die is thrown twice. Let A be the event of getting a prime number when the die is thrown first time and \(B\) be the event of getting an even number when the die is thrown second time. Then \(\mathrm{P}(\mathrm{A} / \overline{\mathrm{B}})=\)

  1. A \(\frac{1}{2}\)
  2. B \(\frac{2}{3}\)
  3. C \(\frac{1}{5}\)
  4. D \(\frac{3}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

\(P(A) = \frac{\text{Number of prime numbers}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}\) \(P(\overline{B}) = \frac{\text{Number of odd numbers}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}\) Events A and B (and thus A and \(\overline{B}\)) are independent.…