Multiple choice

Four fair dice $D_{1},\ D_{2},\ D_{3}$ and $D_{4}$, each having six faces numbered $1, 2, 3, 4, 5$ and $6,$ are rolled simultaneously. The probability that $D_{4}$ shows a number appearing on one of $D_{1},\ D_{2}$ and $D_{3}$ is

  1. $\displaystyle \frac{91}{216}$
  2. $\displaystyle \frac{108}{216}$
  3. $\displaystyle \frac{125}{216}$
  4. $\displaystyle \frac{127}{216}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability that D4 does NOT show a number appearing on D1, D2, or D3 is (5/6)^3 = 125/216. The probability that it DOES show a number appearing on one of the others is 1 - 125/216 = 91/216.

AI explanation

First find the probability that D4 does not match any of D1, D2, or D3. For any value shown on D4, the probability that D1 does not match is 5/6, and the same 5/6 probability applies independently to D2 and D3. The probability of no matches is (5/6) times (5/6) times (5/6), which equals 125/216. Therefore, the probability that D4 shows a number appearing on at least one of the other three dice is 1 minus 125/216, which equals 91/216.

Keep practicing — related questions