# Moderate Probability Solved QuestionAptitude Discussion

 Q. There are three events $A, B$ and $C$, one of which must and only can happen. If the odds are $8:3$ against $A, 5:2$ against $B$, the odds against $C$ must be:
 ✖ A. $13:7$ ✖ B. $3:2$ ✔ C. $43:34$ ✖ D. $43:77$

Solution:
Option(C) is correct

According to the question,

$\dfrac{P(A')}{P(A)}=\dfrac{8}{3},\;P(A)=\dfrac{3}{11}\text{ and }P(A')=\dfrac{8}{11}$

Also $\dfrac{P(B')}{P(B)}=\dfrac{5}{2}$

$P(B)=\dfrac{2}{7}\text{ and }P(B')=\dfrac{5}{7}$

Now, out of A,B and C, one and only one can happen.

$P(A)+P(B)+P(C)=1$

$P(C)=\dfrac{34}{77}$

$P(C')=1-P(C)$

$=\dfrac{43}{77}$

So odd against $C$

$\dfrac{P(C)}{P(C')}=\dfrac{43}{34}$