One of two events must occur. If chance of one is frac{{2}}{{3}} of other, then odds in favour

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14.Probability

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One of the two events must occur. If the chance of one is $\frac{{2}}{{3}}$ of the other, then odds in favour of the other are

A

$2:3$

B

$1:3$

C

$3:1$

D

$3:2$

Solution

(d) Let $p$ be the probability of the other event, then the probability of the first event is $\frac{2}{3}p.$

Since two events are toally exclusive, we have $p + \left( {\frac{2}{3}} \right){\rm{ }}p = 1 \Rightarrow p = \frac{3}{5}$

Hence odds in favour of the other are $3:5 – 3,$ $i.e.$ $3:2$.

Standard 11

Mathematics

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