Multiple choice

The chance of one event happening is the square of the chance of a $2^{nd}$ event, but odds against the first are the cubes of the odds against the 2nd. Find the chances of first event. (Assume that both events are neither sure nor impossible)

  1. $\displaystyle \frac {1}{9}$
  2. $ \dfrac {1}{3}$
  3. $ \dfrac {2}{3}$
  4. $ \dfrac {2}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let p1, p2 be probabilities. p1 = p2^2. Odds against 1 = (1-p1)/p1. Odds against 2 = (1-p2)/p2. (1-p1)/p1 = ((1-p2)/p2)^3. Substituting p1 = p2^2: (1-p2^2)/p2^2 = (1-p2)^3/p2^3. (1-p2)(1+p2)/p2^2 = (1-p2)^3/p2^3. (1+p2)/1 = (1-p2)^2/p2. p2^2 + p2 = 1 - 2p2 + p2^2. 3p2 = 1, p2 = 1/3. p1 = (1/3)^2 = 1/9.

AI explanation

Let the chances of the first and second events be p and q respectively. Given that p = q^2, and the odds against the first are the cubes of the odds against the second, we write (1-p)/p = ((1-q)/q)^3. Substituting q^2 for p gives (1-q^2)/q^2 = ((1-q)/q)^3, which simplifies to (1+q)q = 1-q, yielding 2q^2 + q - 1 = 0. Solving this quadratic equation gives q = 1/2, so the chance of the first event is p = (1/2)^2 = 1/9.

Keep practicing — related questions