box contains 4 chocolates and 1 fruit chew. Clark and Chloe take turns drawing a treat out of the box without replacement. Whoever draws the fruit chew wins. Clark draws first. What is the probability that Chloe wins? Express your answer as a common fraction.

There are five draws in total. If the items that were drawn are listed, there would be five elements in that list in total. Since Clark starts first, the draws of Clark would be the odd-numbered ones: [tex]1,\, 3,\, 5[/tex], while those of Chloe would be the even-numbered ones: [tex]2,\, 4[/tex].

Since the items are taken without replacement, Chloe would have drawn the fruit chew first if and only if one the even-numbered draws is a fruit chew:

[tex]\texttt{C F C C C}[/tex], or
[tex]\texttt{C C C F C}[/tex].

(Chocolate: "[tex]\texttt{C}[/tex]"; Fruit chew: "[tex]\texttt{F}[/tex]".)

Apply the following steps to find the probability that one of the two above situations happen:

Find the number of unique ways to arrange the five items, regardless of whether Chloe won.
Find the number of unique draws in which Chloe won.
Since each of these draws is equally likely, divide the number of unique draws where Chloe won by the total number of draws to find the probability that Chloe would win.

The permutation formula gives the number of ways to arrange [tex]k[/tex] out of a total of [tex]n[/tex] distinct items without replacement:

[tex]\displaystyle P(n,\, k) = \frac{n!}{(n - k)!}[/tex].

In this question, there are a total of [tex]n = 5[/tex] items, and all [tex]k = 5[/tex] items are taken. Hence, the total number of orderings that are possible would be:

[tex]\displaystyle P(5,\, 5) = \frac{5!}{(5 - 5)!} = 5![/tex].

(Note that [tex]0! = 1[/tex].)

In other words, there are a total of [tex]5![/tex] possible ways to draw these items.

To find the number of ways where the fruit chew is drawn in an even position, [tex]\texttt{C F C C C}[/tex] or [tex]\texttt{C C C F C}[/tex], consider each case separately.

For example, in [tex]\texttt{C F C C C}[/tex]:

The first draw is be a chocolate. Initially, there are [tex]4[/tex] possible choices for this chocolate.
The second draw should be a fruit chew, for which there is only [tex]1[/tex] possible choice.
The third draw would be another chocolate. After taking one chocolate in the first draw, there would be [tex]3[/tex] possible options for this chocolate.
Likewise, there would be [tex]2[/tex] possible options for the chocolate for the next draw.
There would be [tex]1[/tex] possible option for the chocolate in the fifth draw.

Hence, the total number of unique draws for [tex]\texttt{C F C C C}[/tex] would be:

[tex]4 \times 1 \times 3 \times 2 \times 1[/tex].

Similarly, there would be [tex]4 \times 1 \times 3 \times 2 \times 1[/tex] unique draws for [tex]\texttt{C C C F C}[/tex].

Since none of the draws of the [tex]\texttt{C F C C C}[/tex] pattern would be counted amongst the draws of the [tex]\texttt{C C C F C}[/tex] pattern, the two sets would be disjoint. The total number of unique draws satisfying either pattern would be the sum of the number for each pattern: [tex]2 \times (4 \times 1 \times 3 \times 2 \times 1)[/tex].

Since each of the [tex]5![/tex] unique draws are equally likely, the probability of reaching a draw where Chloe wins (satisfies either [tex]\texttt{C F C C C}[/tex] or [tex]\texttt{C C C F C}[/tex]) would be equal to the ratio between:

the number of unique draws satisfying the patterns, and
the total number of unique draws.

[tex]\displaystyle \frac{2 \times (4 \times 1 \times 3 \times 2 \times 1)}{5!} = \frac{2}{5}[/tex].

In other words, the probability that Chloe wins would be [tex](2/5)[/tex].