Question

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

(a) 43 : 34

(b) 34 : 43

(c) 43 : 77

(d) 77 : 43

Answer 1

Correct option is (a) 43 : 34

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) = \(\frac 2{5+2} = \frac 27\)

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

⇒ \(\frac 3{11} + \frac 27 + P(C) = 1\)

⇒ \(P(C) = 1 - \frac 3{11} - \frac 27 = \frac{34}{77}\)

Hence, odds against the events are

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

= (77 - 34) : 34

= 43 : 34