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1. The odds against an event are 5 : 3 and the odds in favour of another independent event are 7 : 5. Find the probability that at least one of the two events will occur.

A. $\frac{69}{96}$

B. $\frac{52}{96}$

C. $\frac{71}{96}$

D. $\frac{13}{96}$

Solution:

Let probability of the first event taking place be A and probability of the second event taking place be B.
Then

\begin{aligned} P (A) = \frac{3}{5 + 3} \\ = \frac{3}{8} \\ P (B) = \frac{7}{7 + 5} \\ = \frac{7}{12} \end{aligned}

The required event can be defined as that A takes place and B does not take place (A or B takes place and A does not take place or A takes place and B takes place.)

= [P (A) {1 - P (B)}] +

[P (B) {1 - P (A)}] +

[P (A) P (B)]

= [\frac{3}{8} \times \frac{5}{12}] +

[\frac{5}{8} \times \frac{7}{12}] +

[\frac{3}{8} \times \frac{7}{12}]

\begin{aligned} = \frac{15 + 35 + 21}{96} \\ = \frac{71}{96} \end{aligned}

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