COMEDK · Maths · 36. Probability
Let \(A\) and \(B\) be two events such that one of the two events must occur. Given that the chance of occurrence of \(A\) is \(\dfrac{2}{3}\) the chance of occurrence of \(B\), then odds in favour of \(B\) is
- A \(2: 5\)
- B \(3: 2\)
- C \(2: 3\)
- D \(3: 5\)
Answer & Solution
Correct Answer
(B) \(3: 2\)
Step-by-step Solution
Detailed explanation
Let \(P(A)\) and \(P(B)\) be the probabilities of events \(A\) and \(B\) respectively. Since one of the two events must occur, \(P(A) + P(B) = 1\). Given \(P(A) = \dfrac{2}{3} P(B)\), substituting this into the sum equation: \(\dfrac{2}{3} P(B) + P(B) = 1\)…
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