A six-sided die, with sides numbered 1 through 6, is rolled three times. What is the probability that the sum of the three rolls is at least 4?
A. \(\dfrac1{216}\)
B. \(\dfrac1{54}\)
C. \(\dfrac12\)
D. \(\dfrac{53}{54}\)
E. \(\dfrac{215}{216}\)
Answer: E
Source: Princeton Review
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A six-sided die, with sides numbered 1 through 6, is rolled three times. What is the probability that the sum of the
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Gmat_mission
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By using property of probability
P(A) = 1-P(not A)
A = when sum of dices is at least 4
not A = sum is less than 4 . The only possible option in sum is 3 which is(1,1,1)
probability of (1,1,1) occurring P(not A) = \frac{1}{6} * \frac{1}{6} * \frac{1}{6} = \frac{1}{216}
P(A) = 1- \frac{1}{216} = \frac{215}{216}
E
P(A) = 1-P(not A)
A = when sum of dices is at least 4
not A = sum is less than 4 . The only possible option in sum is 3 which is(1,1,1)
probability of (1,1,1) occurring P(not A) = \frac{1}{6} * \frac{1}{6} * \frac{1}{6} = \frac{1}{216}
P(A) = 1- \frac{1}{216} = \frac{215}{216}
E
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Solution:Gmat_mission wrote: ↑Thu Sep 10, 2020 12:54 amA six-sided die, with sides numbered 1 through 6, is rolled three times. What is the probability that the sum of the three rolls is at least 4?
A. \(\dfrac1{216}\)
B. \(\dfrac1{54}\)
C. \(\dfrac12\)
D. \(\dfrac{53}{54}\)
E. \(\dfrac{215}{216}\)
Answer: E
There are a total of 6 x 6 x 6 = 216 possible combinations when 3 dice are rolled. The only way that the sum is not at least 4 is if all 3 numbers rolled are 1 (notice that 1 + 1 + 1 = 3). Therefore, the probability that the sum is not at least 4 is 1/216, and thus, the probability that the sum is at least 4 is 1 - 1/216 = 215/216.
Answer: E
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