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16 May 2025, 00:30
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C
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Difficulty:
35%
(medium)
Question Stats:
78% (01:24) correct
22%
(01:39)
wrong
based on 69
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Exactly two sides of a certain 10-sided die are red. What is the probability that Kumar rolls the die 3 times and the die lands with a red side up for the first time on the third roll?
(A) 0.032
(B) 0.064
(C) 0.128
(D) 0.200
(E) 0.250
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(A) 0.032
(B) 0.064
(C) 0.128
(D) 0.200
(E) 0.250
GMAT Hacks
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
16 May 2025, 00:46
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We require the probability that a 10-sided die with 2 red sides lands red for the first time on the third roll, rechecking the calculation.
Probability of red is \(\frac{2}{10}\), non-red is \(\frac{8}{10}\). We need: non-red, non-red, red.
Rolls are independent, so multiply:
\(P = \frac{8}{10} \cdot \frac{8}{10} \cdot \frac{2}{10} = \frac{128}{1000} = 0.128\)
Answer: C
Probability of red is \(\frac{2}{10}\), non-red is \(\frac{8}{10}\). We need: non-red, non-red, red.
Rolls are independent, so multiply:
\(P = \frac{8}{10} \cdot \frac{8}{10} \cdot \frac{2}{10} = \frac{128}{1000} = 0.128\)
Answer: C
| Theory Involved |
| Independent Events: For independent events A, B, C (one outcome doesn’t affect others), the probability of all occurring is \(P(A) \cdot P(B) \cdot P(C)\). Each die roll is independent, so we multiply probabilities of non-red, non-red, and red. |
16 May 2025, 05:11
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P(red) = 2/10 = 1/5
P(not red) = 8/10 = 4/5
P(red first time on the third roll) = 4/5 * 4/5 * 1/5 = 16/125 = 0.128
Answer C
P(not red) = 8/10 = 4/5
P(red first time on the third roll) = 4/5 * 4/5 * 1/5 = 16/125 = 0.128
Answer C
Bunuel
