A fair die is thrown three times. If it is known that the three numbers obtained are all different, then the probability that they appear in increasing order (first number \(\lt\) second number \(\lt\) third number) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{6}\)
The \(6^3 = 216\) outcomes are equally likely. The outcomes with three different numbers are
\(6\times 5\times 4 = 120\)
This is the reduced sample space. Any 3 different numbers can be written in increasing order in exactly one way, so the outcomes in increasing order are as many as the ways of choosing 3 numbers out of 6:
\( \displaystyle {}^{6}C_{3} = 20 \)
So
\( \displaystyle P = \frac{20}{120} = \frac{1}{6} \)
In other words, the 3! = 6 orders of the same three numbers are equally likely, and only one of them is increasing.
Why the other options are incorrect
Option (1) \(\displaystyle\frac{5}{54}\) — Incorrect. This is \(\frac{20}{216}\), the probability of increasing order without the given condition.
Option (2) \(\displaystyle\frac{1}{3}\) — Incorrect. This is \(\frac{40}{120}\), the probability that the numbers are in increasing or in decreasing order.
Option (4) \(\displaystyle\frac{5}{9}\) — Incorrect. This is \(\frac{120}{216}\), the probability that the three numbers are all different, which is the condition itself.