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. 

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. 

This implies 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 = \(\frac{8}{3}\) 

⇒ \(\frac{P{(\bar A)}}{P(A)} = \frac{8}{3}\)

⇒ \(\frac{1-P( A)}{P(A)} = \frac{8}{3}\)

⇒ 8 P(A) = 3 – 3 P(A) 

⇒ 11 P(A) = 3

∴ P(A) = \(\frac{3}{11}\) …Equation 2 

Similarly, we are given with: Odd against B = \(\frac{5}{2}\) 

⇒ \(\frac{P{(\bar B)}}{P(B)} = \frac{5}{2}\)

⇒ \(\frac{1-P( B)}{P(B)} = \frac{5}{2}\) 

⇒ 5 P(B) = 2 – 2 P(B) 

⇒ 7 P(B) = 2 

∴ P(B) = \(\frac{2}{7}\)…Equation 3

From equation 1,2 and 3 we get:

P(C) = 1- \(\frac{3}{11}-\frac{2}{7}=\frac{77-21-22}{77} = \frac{34}{77}\)  

∴ P(C’) = 1 – \(\Big(\frac{34}{77}\Big)=\frac{43}{77}\)  

∴ Odd against C = \(\frac{p{(\bar c)}}{p(c)}\) = \(\frac{43}{\frac{77}{\frac{34}{77}}}\) = \(\frac{43}{34}\)