ProbabilityHardBloom L3
Question
An unbiased die with faces marked 1, 2, 3, 4, 5, and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5 is
Options
A.$\frac{16}{81}$
B.$\frac{1}{81}$
C.$\frac{80}{81}$
D.$\frac{65}{81}$
Solution
{"given":"An unbiased die with faces 1, 2, 3, 4, 5, 6 is rolled four times. We need to find the probability that all four outcomes have minimum value ≥ 2 and maximum value ≤ 5.","key_observation":"For the condition to be satisfied, each of the four rolls must result in a face value from the set {2, 3, 4, 5}. Since the rolls are independent events, we can use the multiplication principle. The favorable outcomes per roll are 4 out of 6 possible outcomes.","option_analysis":[{"label":"(A)","text":"$\\frac{16}{81}$","verdict":"correct","explanation":"Each roll must be from {2,3,4,5}, giving probability $\\frac{4}{6} = \\frac{2}{3}$ per roll. For four independent rolls: $\\left(\\frac{2}{3}\\right)^4 = \\frac{16}{81}$."},{"label":"(B)","text":"$\\frac{1}{81}$","verdict":"incorrect","explanation":"This would correspond to $\\left(\\frac{1}{3}\\right)^4$, which is too restrictive. This might arise from considering only one specific outcome like all 3's."},{"label":"(C)","text":"$\\frac{80}{81}$","verdict":"incorrect","explanation":"This is $1 - \\frac{1}{81}$ and represents the complement of option B, not the correct probability for our constraint."},{"label":"(D)","text":"$\\frac{65}{81}$","verdict":"incorrect","explanation":"This value doesn't correspond to any logical combination of probabilities for this problem setup and appears to be a distractor."}],"answer":"(A)","formula_steps":[]}
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