Multiple choice

The chance of an event happening is the square of the chance, of a second event but the odds against the first are the cubes of the odds against the first are the cubes of the odds against the second. Find the chance of each.

  1. probability of first event $\displaystyle =\dfrac{1}{9}$ ,probability of second event $\displaystyle =\dfrac{1}{3}$
  2. probability of first event $\displaystyle =\dfrac{1}{3}$, probability of second event $\displaystyle =\dfrac{1}{9}$
  3. probability of first event $\displaystyle =\dfrac{1}{4}$, probability of second event $\displaystyle =\dfrac{1}{2}$
  4. probability of first event $\displaystyle =\dfrac{1}{2}$, probability of second event $\displaystyle =\dfrac{1}{4}$
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A Correct answer
Explanation

Let the second probability be q, so the first probability is q^2. Using odds against, (1 - q^2)/q^2 = ((1 - q)/q)^3, which gives q = 1/3. Hence the first probability is 1/9 and the second is 1/3.

AI explanation

Let p be the probability of the first event and q be the probability of the second event. Based on the problem statement, we have p equals q squared. The odds against the first and second events are (1-p) divided by p and (1-q) divided by q, respectively, leading to the equation (1-p) divided by p equals ((1-q) divided by q) cubed. Substituting p equals q squared into the second equation and solving yields q equals 1 divided by 3. Substituting this back into the first equation gives p equals 1 divided by 9, making the probability of the first event 1/9 and the second event 1/3.

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