Three numbers a, b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that $a^{3} + b^{3} + c^{3} - 3 ab c$ is divisible by 3 is

Answer (Detailed Solution Below)View answer ▾

Answer: Option A — $\frac{3 \cdot ^{33} C _{3} + ( ^{33} C _{1} ) ^{3}}{ ^{99} C _{3}}$

Detailed Solution

$n^{3} \equiv n (m o d 3)$ for every integer n, so $a^{3} + b^{3} + c^{3} - 3 ab c \equiv a + b + c (m o d 3)$ .

The numbers 1 to 99 fall into three remainder classes (0, 1, 2 mod 3) of 33 numbers each. $a + b + c \equiv 0$ exactly when all three come from one class ( $3 \cdot ^{33} C_{3}$ ways) or one comes from each class ( $(^{33} C_{1})^{3}$ ways).

$P = \frac{3 \cdot ^{33} C _{3} + ( ^{33} C _{1} ) ^{3}}{ ^{99} C _{3}}$ .

✓ Check

Same count on 1 to 6 (two per class):

3 \cdot ​^{2} C_{3} + 2^{3} = 8

of the

​^{6} C_{3} = 20

triples, and listing them gives {1,2,3}, {1,2,6}, {1,3,5}, {1,5,6}, {2,3,4}, {2,4,6}, {3,4,5}, {4,5,6}: 8.

⚠ Common trap

2 \cdot ​^{33} C_{3}

); the class of multiples of 3 also works.

Quick tip

$n^{3} \equiv n (mod 3) \Rightarrow$ need $a + b + c \equiv 0$ : all same class or one from each class

💬 Discussions on this question

Ask a doubt ›

Still stuck on this question? Ask our faculty and fellow students — answers appear here.

Ask a doubt

Probability — Preparation

This question tests your understanding of Probability, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Probability measures how likely an event is, as a number between 0 and 1. It equals the number of favourable outcomes divided by the total number of possible outcomes, when all outcomes are equally likely. This topic covers coins, dice, playing cards, drawing balls from a bag, complementary events, independent events, and conditional probability. The most useful tools are the complement rule (P(at least one) = 1 - P(none)) and the multiplication rule for independent events. For engineering and MCA entrance exams, the topic also includes the addition theorem, conditional probability, Bayes' theorem, and probability distributions. A clear sample space is the key to avoiding mistakes, so always list or count all outcomes carefully. Practise the probability questions below. Each comes with a step-by-step solution that shows how the sample space and favourable outcomes are counted.

‹ All Probability questions Browse all topics ›

Three numbers a, b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that $a^{3} + b^{3} + c^{3} - 3 ab c$ is divisible by 3 is

💬 Discussions on this question

More Probability Questions

More Quantitative Aptitude Questions

Probability — Preparation

Three numbers a, b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that a3+b3+c3−3abc is divisible by 3 is

💬 Discussions on this question

More Probability Questions

More Quantitative Aptitude Questions

Probability — Preparation

Three numbers a, b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that $a^{3} + b^{3} + c^{3} - 3 ab c$ is divisible by 3 is