Chapter 8, Problem 77CR

To determine

To find: the probability that each side appears exactly once.

Answer to Problem 77CR

The probability that each side appears exactly once = $\frac{24}{625}$ .

Explanation of Solution

Given information: A five-sided die is rolled five times.

Formula used:

Formula for the number of combinations of n numbers taken r numbers at a time is,

  $n(S){=}_n{C}_r.$

If an event has n(E) equally likely outcomes and its sample space S has n(S) equally likely outcomes, then the probability of event is given by,

  $P(E)=\frac{n(E)}{n(S)}.$

Calculation:

On the first roll, there are no restrictions. The die is allowed to be any of the 5 equally likely values. Thus the probability of not duplicating any numbers so far after roll 1 is, $\frac{5}{5}.$

For each subsequent roll, the number of "successful" rolls decreases by 1. For instance, if our first roll was a [3], then the second roll needs to be anything but [3], meaning there are 4 "successful" outcomes (out of the 5 possible) for roll 2. So, since each roll is independent of the previous rolls, multiply their "success" probabilities together. The probability of rolling no repeats after two rolls is $\frac{5}{5}\times \frac{4}{5}$ .

Continuing this pattern, the third roll will have 3 "successful" outcomes out of 5 possible, so, The probability of rolling no repeats after three rolls is $\frac{5}{5}\times \frac{4}{5}\times \frac{3}{5}$ .

The fourth roll will have 2 "successful" outcomes out of 5 possible, so, The probability of rolling no repeats after fourth rolls is $\frac{5}{5}\times \frac{4}{5}\times \frac{3}{5}\times \frac{2}{5}$ .

The fifth roll will have 1 "successful" outcome out of 5 possible, so, the probability of rolling no repeats after fourth rolls is $\frac{5}{5}\times \frac{4}{5}\times \frac{3}{5}\times \frac{2}{5}\times \frac{1}{5}=\frac{24}{625}.$