Question

There are three events X, Y and Z, out of which only one must happen. If the odds are 8:3 against X, 5:2 against Y, the odds against Z must be:

• A. 43:34
• B. 8:11
• C. 7:5
• D. None of these

Answer 1

The correct option is A

43:34

According to the question,$\frac{P\left(X^{\prime }\right)}{P\left(X\right)}$ = $\frac{8}{3}$, P(X') = $\frac{8}{11}$, P(X) = $\frac{3}{11}$

Similarly P(Y') = $\frac{5}{7}$, P(Y) = $\frac{2}{7}$

Now, out of X,Y and Z, one and only one can happen.

P(X) + P(Y) + P(Z) = 1

So, P(Z) = $\frac{34}{77}$

P(Z${}^{\prime }$) = 1-P(Z) = $\frac{43}{77}$

So, odds against Z = 43:34.