Probability
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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.

$43:77$

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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 $C$

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


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