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. $$\frac{{13}}{{96}}$$
Answer: Option C
Solution(By Examveda Team)
Let probability of the first event taking place be A and probability of the second event taking place be B.Then
$$\eqalign{ & P\left( A \right) = \frac{3}{{5 + 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{8} \cr & P\left( B \right) = \frac{7}{{7 + 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{7}{{12}} \cr} $$
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.)
$$ = \left[ {P\left( A \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$ $$\left[ {P\left( B \right)\left\{ {1 - P\left( A \right)} \right\}} \right] + $$ $$\left[ {P\left( A \right)P\left( B \right)} \right]$$
$$ = \left[ {\frac{3}{8} \times \frac{5}{{12}}} \right] + $$ $$\left[ {\frac{5}{8} \times \frac{7}{{12}}} \right] + $$ $$\left[ {\frac{3}{8} \times \frac{7}{{12}}} \right]$$
$$\eqalign{ & = \frac{{15 + 35 + 21}}{{96}} \cr & = \frac{{71}}{{96}} \cr} $$
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
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