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One of the two events must occur. If the chance of one is 2/3 of the other, then odds in favour of the other are A. 1 : 3 B. 3 : 1 C. 2 : 3 D. 3 : 2 |
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Answer» Let E and F be the two events such that one must occur Given, P(E) = \(\frac{2}{3}\) P(F) Also, P(E∪F) = 1 P(E) + P(F) = 1 ⇒ P(F){ \(\frac{2}{3}\) + 1} = 1 ∴ P(F) = \(\frac{3}{5}\) And P(F’) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\) We have to find \(\frac{P(F)}{P(\bar F)}\) = \(\frac{3}{\frac{5}{\frac{2}{5}}}\) = \(\frac{3}{2}\) ∴ Odds in favour of F = \(\frac{3}{2}\) As our answer matches only with option (d) ∴ Option (d) is the only correct choice. |
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