The chance of an event happening is the square of the chance of a second event but the odds against the first are the cube of the odds against the second. The chances of the events are

Options

A. `P_(1) = 1//9`
B. `P_(1) = 1//16`
C. `P_(2) = 1//3`
D. `P_(2) = 1//4`

Answer

A, C

Explanation

Let `p_(1), p_(2)` be the chances of happening of the first and second events, respectively. Then according to the given conditions, we have
`p_(1) = p_(2)^(2)`
and `(1-p_(1))/(p_(1)) = ((1-p_(2))/(p_(2)))^(3)`
Hence,
`(1-p_(2)^(2))/(p_(2)^(2)) = ((1-p_(2))/(p_(2)))^(3) or p_(2)(1 + p_(2)) = (1 - p_(2))^(2)`
or `3p_(2) = 1 or p_(2) = (1)/(3) " and so "p_(1) = (1)/(9)`