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&nbsp;a<sup>3</sup> + b<sup>3</sup> + c<sup>3</sup> - 3abc&nbsp;is divisible by 3, is

(A) <span class="math-tex">\(\rm\frac{3\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)</span>
(B) <span class="math-tex">\(\rm\frac{3\times {^{33}C_3}- \left(^{33}C_1\right)^3}{^{99}C_3}\)</span>
(C) <span class="math-tex">\(\rm\frac{2\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)</span>
(D) <span class="math-tex">\(\rm \frac{2\times {^{33}C_3}-\left(^{33}C_1\right)^3}{^{99}C_3}\)</span>
View Dynamic Solution & Explanation
Correct Solution: Option A

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