Chapter 20, Problem 8E

(a)

To determine

To find out what is the probability that you would have picked three vanillas in a row.

Answer to Problem 8E

  $0.0909$ .

Explanation of Solution

It is given in the question that there are six vanilla and six peanut butter candies in the box. Thus, we then have six possible vanilla candies out of $12$ candies. So, we have,

  $\begin{array}{l}P(\text{First vanilla})=\frac{6}{12}\\ =\frac{1}{2}\end{array}$

After the first vanilla candy is selected, there are five vanilla candies left out of $11$ candies. So, we have,

  $P(\text{Second vanilla | First vanilla})=\frac{5}{11}$

After the first and second vanilla candy are selected there are four vanilla candies left out of ten candies. So, we have,

  $\begin{array}{l}P(\text{Third vanilla | First and second vanilla})=\frac{4}{10}\\ =\frac{2}{5}\end{array}$

Thus, now multiply the probabilities of each case above as:

  $\begin{array}{l}P(\text{Three vanilla candies})=\frac{1}{2}\times \frac{5}{11}\times \frac{2}{5}\\ =\frac{1}{11}\\ =0.0909\end{array}$

(b)

To determine

To explain do you think three really might have been six of each.

Answer to Problem 8E

Yes, we think that three really might have been six of each.

Explanation of Solution

It is given in the question that there are six vanilla and six peanut butter candies in the box. Thus, we then have six possible vanilla candies out of $12$ candies. So, we have,

  $\begin{array}{l}P(\text{Three vanilla candies})=\frac{1}{2}\times \frac{5}{11}\times \frac{2}{5}\\ =\frac{1}{11}\\ =0.0909\end{array}$

Thus, the probability that you would have picked three vanillas in a rowis $0.0909$ . So, we can say that we think three really might have been six of each because the probability in part (a) is not small as it is larger than $0.05$ and thus it appears to there really might have been six of each.

(c)

To determine

To explain would you continue to believe that half are vanilla if the fourth one you try is also vanilla.

Answer to Problem 8E

No, we would not continue to believe that half are vanilla if the fourth one you try is also vanilla.

Explanation of Solution

It is given in the question that there are six vanilla and six peanut butter candies in the box. Thus, we then have six possible vanilla candies out of $12$ candies. So, we have,

  $\begin{array}{l}P(\text{First vanilla})=\frac{6}{12}\\ =\frac{1}{2}\end{array}$

After the first vanilla candy is selected, there are five vanilla candies left out of $11$ candies. So, we have,

  $P(\text{Second vanilla | First vanilla})=\frac{5}{11}$

After the first and second vanilla candy are selected there are four vanilla candies left out of ten candies. So, we have,

  $\begin{array}{l}P(\text{Third vanilla | First and second vanilla})=\frac{4}{10}\\ =\frac{2}{5}\end{array}$

After the first, second and third vanilla candy are selected there are three vanilla candies left out of nine candies. So, we have,

  $\begin{array}{l}P(\text{Fourth vanilla | First, second and third vanilla})=\frac{3}{9}\\ =\frac{1}{3}\end{array}$

Thus, now multiply the probabilities above as:

  $\begin{array}{l}P(\text{Four vanilla candies})=\frac{1}{2}\times \frac{5}{11}\times \frac{2}{5}\times \frac{1}{3}\\ =\frac{1}{33}\\ =0.0303\end{array}$

Since the probability is small then it appears that more than half are vanilla. No, we would not continue to believe that half are vanilla if the fourth one you try is also vanilla.