One of the two events must happen. Given that the chance of one is two

1.	One of the two events must happen. Given that the chance of one is two-third of the other, find the odds in favour of the other.
Answer» Let A and B are two events. As, out of 2 events A and B only one can happen at a time which means no event have anything common. ∴ We can say that A and B are mutually exclusive events. By definition of mutually exclusive events we know that: P(A ∪ B) = P(A) + P(B) According to question one event must happen. This implies A or B is a sure event. ∴ P(A ∪ B) = P(A) + P(B) = 1 …Equation 1 Given, P(A) = (2/3)P(B) To find: odds in favour of B ∴ P(B’) + \(\frac{2}{3}\) P(B) = 1 ⇒ \(\frac{5}{3}\) P(B) = 1 ⇒ P(B) = \(\frac{3}{5}\) ∴ P(B’) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\) ∴ Odd in favour of B = \(\frac{P(B)}{P(\bar B)}\) = \(\frac{3}{\frac{5}{\frac{2}{5}}}\) = \(\frac{3}{2}\)

One of the two events must happen. Given that the chance of one is two-third of the other, find the odds in favour of the other.

Answer»

Let A and B are two events.

As, out of 2 events A and B only one can happen at a time which means no event have anything common.

∴ We can say that A and B are mutually exclusive events.

By definition of mutually exclusive events we know that:

P(A ∪ B) = P(A) + P(B)

According to question one event must happen.

This implies A or B is a sure event.

∴ P(A ∪ B) = P(A) + P(B) = 1 …Equation 1

Given, P(A) = (2/3)P(B)

To find: odds in favour of B

∴ P(B’) + \(\frac{2}{3}\) P(B) = 1

⇒ \(\frac{5}{3}\) P(B) = 1 ⇒ P(B) = \(\frac{3}{5}\)

∴ P(B’) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)

∴ Odd in favour of B = \(\frac{P(B)}{P(\bar B)}\) = \(\frac{3}{\frac{5}{\frac{2}{5}}}\) = \(\frac{3}{2}\)

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