Chapter 10, Problem 5RE

Solution

To determine: The probability that a person who starts eating them one at a time won’t get a caramel until the third candy.

The probability that a person who starts eating them one at a time won’t get a caramel until the third candy.is 0.033.

Given information:

The box of candies contains 6 caramels and 4 buttercreams.

Formula used:

  $\mathrm{P}=\frac{E}{S}$   ...... (1)

Here, E is the favourable outcomes and S is the total possible outcomes.

And,

Use general multiplication rule:

P (A and B) = P(A)×P(B|A) = P(B)×P(A|B)    ...... (2)

Calculation:

When the first caramel is obtained on the third candy, then the first two candies were buttercreams and the third candy was a caramel.

Let

  $B_1$ =First draw results in buttercream

  $B_2$ =Second draw results in buttercream

  $C_3$ =Third draw results in caramel.

4 of the 4+6 = 10 candies in the box are butter creams.

Substitute 4 for $E$ and 10 for $S$ in equation (1)

  $\mathrm{P}(B_1)=\frac{4}{10}$

When the first candy was a buttercream, then 3 of the 9 remaining candies in the box are butter creams.

Substitute 3 for $E$ and 9 for $S$ in equation (1)

  $\mathrm{P}(B_2|B_1)=\frac{3}{9}$

When the first two candies were a buttercream, then 6 of the 8 remaining candies in the box are caramels.

Substitute 2 for $E$ and 8 for $S$ in equation (1)

  $\mathrm{P}(C_3|B_1{B}_2)=\frac{2}{8}$

Substitute $\frac{4}{10}$ for $\mathrm{P}(B_1)$ , $\frac{3}{9}$ for $\mathrm{P}(B_2|B_1)$ and $\frac{2}{8}$ for $\mathrm{P}(C_3|B_1{B}_2)$ in equation (2)

  $\begin{array}{c}P\left(B_1{B}_2{C}_3\right)=\frac{4}{10}\times \frac{3}{9}\times \frac{2}{8}\\ =\frac{24}{720}\\ =\frac{1}{30}\\ =0.033\end{array}$

Hence, the probability that a person who starts eating them one at a time won’t get a caramel until the third candy.is 0.033.