Chapter 10.1, Problem 48E

Solution

To calculate: The probability that you have to buy 5 of them if it is need to buy gumballs until it will get a blue one.

The probability it has to buy $5$ of them is $39.91\%$

Given information:

A vending machine contains $12$ red gumballs, $5$ white ones, and $3$ blue ones. You insert 3 coins, getting one gumball after another after another. It is need to buy gumballs until you get a blue one.

Calculation:

Let $A_i=$ blue gumball on  $i\text{th}$ draw

  $A_i^c=$ not a blue gumball on  $i\text{th}$ draw

  $i=1,2,\mathrm{...},5$

  $P(A)=40\%=0.4$

First, buy $5$ gumballs when the first four gumballs are all not blue and the fifth gumball is blue.

Initially, $3$ of the  $12+5+3=20$  balls in the vending machine are blue and thus  $20-3=17$  of the $20$ balls are not blue.

The probability is the number of favorable outcomes divided by the number of possible outcomes:

  $\begin{array}{l}P(A_1^c\text{})=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}\\ =\text{}\frac{17}{20}\end{array}$

After a non-blue gumball has been selected, 16 of the remaining 19 balls are not blue.

  $\begin{array}{l}P(A_2^c|A_1^c\text{})=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}\\ =\text{}\frac{16}{19}\end{array}$

After two non-blue gumballs have been selected, 15 of the remaining 18 balls are not blue.

  $\begin{array}{l}P(A_3^c|A_1^c{A}_2^c\text{})=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}\\ =\text{}\frac{15}{18}\end{array}$

After three non-blue gumballs have been selected, 14 of the remaining 17 balls are not blue.

  $\begin{array}{l}P(A_4^c|A_1^c{A}_2^c{A}_3^c\text{})=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}\\ =\text{}\frac{14}{17}\end{array}$

After four non-blue gumballs have been selected, 3 of the remaining 16 balls are blue.

  $\begin{array}{l}P(A_5|A_1^c{A}_2^c\text{}{A}_3^c{A}_4^c)=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}\\ =\text{}\frac{13}{16}\end{array}$

Use the general multiplication rule: $P(A\text{ and }B)=P(A)\times P(B|A)=P(B)\times P(A|B)$

  $\begin{array}{l}P(A_1^c{A}_2^c\text{}{A}_3^c{A}_4^c{A}_5)=P\left(A_1^c\right)\times P\left(A_2^c\right)\times P\left(A_3^c\right)\times P\left(A_4^c\right)\times P\left(A_5\right)\\ =\frac{17}{20}\times \frac{16}{19}\times \frac{15}{18}\times \frac{14}{17}\times \frac{13}{16}\\ =\frac{91}{228}\\ =0.3991\\ \approx 39.91\%\end{array}$

Hence, the probability that it has to buy 5 of them if it is need to buy gumballs until it will get a blue one is $39.91\%$ .