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.

Asked by Aaryan | 1 year ago |  36

##### Solution :-

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.

So, by definition of mutually exclusive events we know that:

P (A ∪ B) = P (A) + P (B)

According to question one event must happen.

A or B is a sure event.

So, P (A ∪ B) = P (A) + P (B) = 1 … Equation (1)

Given: P (A) = ($$\dfrac{2}{3}$$) P (B)

We have to find the odds in favour of B.

P (B) = $$\dfrac{3}{5}$$

So, P (B′) = 1 – $$\dfrac{3}{5}$$

=$$\dfrac{2}{5}$$

Odd in favour of B: $$\dfrac{3}{2}$$

Answered by Aaryan | 1 year ago

### Related Questions

#### One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

#### The probability that a student will pass the final examination in both English and Hindi is 0.5

The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75. What is the probability of passing the Hindi examination?

#### A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.

A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.