Question

Box of chocolates According to Forrest Gump, “Life is like a box of chocolates. You never know what you’re gonna get.” Suppose a candy maker offers a

special “Gump box” with $20$ chocolate candies that look the same. In fact, $14$ of the candies have soft centers and 6 have hard centers. Choose $2$ of the

candies from a Gump box at random.

(a) Draw a tree diagram that shows the sample space of this chance process.

(b) Find the probability that one of the chocolates has a soft center and the other one doesn’t.

Short answer

Part (b) P (Hard center after Soft center) = $0.4421$

Part (a) The tree diagram is

Step-by-step solution

Part (a) Step 1. Given Information

A candy company sells a special "Gump box" that contains $20$ chocolates, $14$ of which have soft centers and 6 of which have hard centers.

Part (a) Step 2. Concept Used

A tree diagram can be used to depict the sample space when chance behavior involves a series of outcomes. Tree diagrams can also be used to determine the likelihood of two or more events occurring at the same time. Simply multiplying along the branches that correspond to the desired results is all that is required.

Part (a) Step 3. Explanation

There are two choices, therefore at each knot, two branches are needed:

The probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes:

Part (b) Step 3. Calculation

$P(HardcenterafterSoftcenter)=0.4421$

Multiplying the related probabilities to determine the likelihood that one of the chocolates has a soft center while the other does not

Therefore, $P\left(SH\right)=P(Hardcenter)+P(SoftcenterafterHardcenter)$

$=\frac{6}{20}\times \frac{14}{19}=\frac{84}{30}$

$P\left(HS\right)=P(Softcenter)+P(HardcenterafterSoftcenter)$

$=\frac{14}{20}\times \frac{6}{19}=\frac{84}{30}$

To find the likelihood that one of the chocolates has a soft center and the other does not add the related probabilities.

Thus, $P(HardcenterafterSoftcenter)=P\left(HS\right)+P\left(SH\right)$

$=\frac{84}{380}+\frac{84}{380}=\frac{168}{380}=0.4421$

As a result, the probability of one of the chocolates having a soft center while the other does not is $P(HardcenterafterSoftcenter)=0.4421$