Question

A symmetric die is rolled 3 times. If it is known that face 1 appeared at least once what is the probability that it appeared exactly once?

Answer 1

The answer is 215/216
That is the probability that it appeared exactly once is 215/216

Answer 2

Answer:

The probability that face 1 appeared exactly once given that face 1 appears at least once is $\frac{75}{91}$.

Step-by-step explanation:

The probability of face 1 occurring exactly once six is

= $\binom{3}{1} (\frac{1}{6})(\frac{5}{6} )(\frac{5}{6} ) = \frac{75}{216}$  ....    (i)

The probability of face 1 not appearing at all

= $(\frac{5}{6} )(\frac{5}{6} )(\frac{5}{6} ) = \frac{125}{216}$ ....    (ii)

The probability of face 1 appearing at least once

=  1  - $\frac{125}{216}$ = $\frac{91}{216}$     ....   (iii)

The probability that face 1 appeared exactly once given that face 1 appears at least once  = $\frac{\frac{75}{216} }{\frac{91}{216} } = \frac{75}{91}$

This is also a conditional probability and hence to get the right answer we have to divide the result from (i) by the result from (iii)

The probability that face 1 appeared exactly once given that face 1 appears at least once is $\frac{75}{91}$.