Multiple choice

An unbiased die with faces $1,2,3,4,5,6$ is tossed $4$ times out of four face values obtained the probability that minimum face value is not less than $2$ and the maximum face value is not greater than $5$ is

  1. $ \displaystyle \frac{1}{81} $
  2. $ \displaystyle \frac{80}{81} $
  3. $ \displaystyle \frac{64}{81} $
  4. $ \displaystyle \frac{16}{81} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Each toss has 6 outcomes. We want faces in {2, 3, 4, 5}. There are 4 such values. Probability for one toss is 4/6 = 2/3. For 4 tosses, the probability that all are in {2, 3, 4, 5} is (2/3)^4 = 16/81.

AI explanation

For the minimum value to not be less than 2 and the maximum to not be greater than 5, every roll must result in either a 2, 3, 4, or 5. This gives 4 acceptable outcomes for each of the 4 throws, resulting in 4^4 = 256 favorable combinations. Since the total possible outcomes for four rolls of a die are 6^4 = 1296, the probability is 256 / 1296, which simplifies to 16 / 81.

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