There are three events A, B, C one of which must and only one can happen, the odds are 8 to 3 against A, 5 to 2 against B, fins the odds against C.

Asked by Aaryan | 1 year ago |  32

##### Solution :-

As, out of 3 events A, B and C only one can happen at a time which means no event have anything common.

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

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

P (A ∪ B ∪ C) = P (A) + P (B) + P (C)

According to question one event must happen.

So, A or B or C is a sure event.

P (A ∪ B ∪ C) = 1 … Equation (1)

We need to find odd against C:

Given:

Odd against A = $$\dfrac{8}{3}$$

8 P (A) = 3 – 3 P (A)

11 P (A) = 3

P (A) = $$\dfrac{3}{11}$$ … Equation (2)

Similarly, we are given with: Odd against B = $$\dfrac{5}{2}$$ 5 P (B) = 2 – 2 P(B)

7 P (B) = 2

P (B) =$$\dfrac{2}{7}$$ …Equation (3)

From equation 1, 2 and 3 we get:

P (C) = 1 – $$\dfrac{3}{11}$$ – $$\dfrac{2}{7}$$

= $$\dfrac{ (77-21-22)}{77}$$

$$\dfrac{34}{77}$$

So, P (C′) = 1 – ($$\dfrac{34}{77}$$)

=$$\dfrac{43}{77}$$

Odd against C:

$$\dfrac{43}{34}$$

Answered by Aaryan | 1 year ago

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