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Question

# The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that (i) at least one of the events will occur, and (ii) none of the events will occur.

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Solution

## $\text{The odds against event A are 5 to 2.}\phantom{\rule{0ex}{0ex}}P\left(A\right)=\frac{2}{5+2}=\frac{2}{7}\phantom{\rule{0ex}{0ex}}\mathrm{The odds}\text{in favour of event B are 6 to 5.}\phantom{\rule{0ex}{0ex}}P\left(B\right)=\frac{6}{6+5}=\frac{6}{11}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\text{P}\left(\text{atleast one event occurs}\right)\phantom{\rule{0ex}{0ex}}=P\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\phantom{\rule{0ex}{0ex}}=P\left(A\right)+P\left(B\right)-P\left(A\right)×P\left(B\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{7}+\frac{6}{11}-\frac{2}{7}×\frac{6}{11}\phantom{\rule{0ex}{0ex}}=\frac{22+42}{77}-\frac{12}{77}\phantom{\rule{0ex}{0ex}}=\frac{22+42-12}{77}=\frac{52}{77}\phantom{\rule{0ex}{0ex}}\therefore P\left(A\cup B\right)=\frac{52}{77}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)P\left(\text{none of the event occurs}\right)\phantom{\rule{0ex}{0ex}}=1-P\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}=1-\frac{52}{77}\phantom{\rule{0ex}{0ex}}=\frac{25}{77}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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