Three numbers a, b and c are chosen at... - Solved

Question 4

Mathematics Probability Hard

Three numbers a, b and c are chosen at random (simultaneously) from among the numbers 1, 2, 3, ..., 99. The probability that a3 + b3 + c3 - 3abc is divisible by 3, is

(A) \(\rm\frac{3\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)

(B) \(\rm\frac{3\times {^{33}C_3}- \left(^{33}C_1\right)^3}{^{99}C_3}\)

(C) \(\rm\frac{2\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)

(D) \(\rm \frac{2\times {^{33}C_3}-\left(^{33}C_1\right)^3}{^{99}C_3}\)

✓ Correct Answer: Option A Verified Solution

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Correct Option is A. Practice regular variations of this Probability concept to build speed and accuracy.

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Question 4

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