Question

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, find the odds against C

Answer 1

Since the odds against event A are 8 : 3, the probability of the happening of event A is given by
P(A) = \[\frac{3}{8 + 3} = \frac{3}{11}\]

Similarly, the odds against event B are 5 : 2.
So, P(B) = \[\frac{2}{5 + 2} = \frac{2}{7}\]

Since events A, B and C are such that one of them must and only one can happen, A, B and C are mutually exclusive and totally exhaustive events.
Consequently, A ∪ B ∪ C = S
and A ∩ B = B ∩ C = C ∩ A = Φ
Thus, P (A ∪ B ∪ C) = P(S) = 1
⇒ P(A) + P(B) + P(C) = 1
⇒\[\frac{3}{11} + \frac{2}{7} + P\left( C \right) = 1\]

\[\Rightarrow P\left( C \right) = 1 - \frac{3}{11} - \frac{2}{7} = \frac{34}{77}\]

Hence, odds against the events are

 

\[P\left( \bar{C} \right): P\left( C \right) = \left( 1 - \frac{34}{77} \right): \frac{34}{77}\] 

                          = (77-34) : 34
                          = 43 : 34