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

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:

\( \dfrac{43}{34}\)