The Relationship Between Independence and Conditional Probability

From the given diagram, the sample space contains $20$ different outcomes. The required probabilities can be found one at a time.

• To find $P(H),$ divide the number of outcomes in the sample space with a History book in the first position by $20.$
• To find $P(G),$ divide the number of outcomes in the sample space with a Geometry book in the second position by $20.$
• To find $P(H\cap G),$ divide the number of outcomes in the sample space with a History book in the first position and a Geometry book in the second position by $20.$

The following diagram summarizes all the computations. Notice that $P(H\cap G)\ne P(H)\cdot P(G).$ That relationship implies that the given events are dependent. Start by finding the ratio of $P(H\cap G)$ to $P(H).$ Next, find the ratio of $P(H\cap G)$ to $P(G).$ As seen, there are $8$ outcomes in the new sample space and $6$ of them have a Geometry book in the second position. Therefore, the probability that the second book is a Geometry book, given that the first book is a History book, can be found using the following process. Consequently, if the first book chosen by Diego is a History book, the probability that the second book is a Geometry book is $\frac{3}{4}.$ As seen, there are $12$ outcomes in the new sample space and $6$ of them have a History book in the first position. Therefore, the probability that the first book is a History book, given that the second book is a Geometry book can be found as follows. Consequently, if the second book chosen by Diego is a Geometry book, the probability that the first book is a History book is $\frac{1}{2}.$ Comparing the four probabilities, it can be seen that the first probability found in part B equals the probability found in part C.

The probability that the second book is a Geometry book given that the first book chosen is a History book equals $\frac{P(H\cap G)}{P(H)}.$

The previous statement can also be rewritten in terms of $H$ and $G$ as follows.

The probability that event $G$ happens given that event $H$ happened equals $\frac{P(H\cap G)}{P(H)}.$

Similarly, the second probability found in part B equals the probability found in part D. This leads to write the following relation.

The probability that the first book is a History book, given that the second book is a Geometry book equals $\frac{P(H\cap G)}{P(G)}.$

As before, the previous statement can be rewritten in terms of $H$ and $G.$

The probability that event $H$ happens, given that event $G$ happened equals $\frac{P(H\cap G)}{P(G)}.$

All of the previous stages and events can be illustrated using a tree diagram. Recall that there are $5000$ bags, and $7\%$ of them contain forbidden items. The product of these numbers will give the number of the bags that contain forbidden items. The rest of the bags do not contain forbidden items. Additionally, the following two details about the bags are known.

• If a bag contains forbidden items, there is a $98\%$ chance that it triggers the alarm.
• If a bag does not have forbidden items, there is a $4\%$ chance that it triggers the alarm.

Considering these details, it can be concluded that $2\%$ of the bags containing forbidden items could trigger the alarm and $96\%$ of the bags that do not have forbidden items could not trigger the alarm.

Now, using the percentages in the branches, the number of bags for each event can be found.

Forbidden and Alarm	$350\cdot 98\%=343$
Forbidden and No Alarm	$350\cdot 2\%=7$
Not Forbidden and Alarm	$4650\cdot 4\%=186$
Not Forbidden and No Alarm	$4650\cdot 96\%=4464$

Finally, all the information can be shown on the tree diagram.

To find the probability that a bag triggers the alarm, start by finding how many bags triggered the alarm. To do so, add the two corresponding numbers in the tree diagram. The probability that a randomly picked bag triggers the alarm $P(\text{Alarm})$ is obtained dividing $529$ by the total number of bags. Since the total number of bags is $5000,$ the ratio of the number of the favorable outcomes to the total number will give the desired probability.

Probabilities of the Events
$P(\text{Alarm and Forbidden})$	$\frac{343}{5000}=6.86\%$
$P(\text{No Alarm and Forbidden})$	$\frac{7}{5000}=0.14\%$

Take note that the sum of the probabilities is equal to $7\%,$ which is the percentage of the bags that contain forbidden items.

As presented in the tree diagram, the number of bags that trigger the alarm is equal to the sum of the numbers under the word Alarm. Therefore, there are $529$ bags that trigger the alarm. Since Mark's bag triggered the alarm, his bag is one of those $529.$ Out of this group of bags, $343$ had forbidden items. Therefore, there is about a $64.84\%$ chance that Mark's bag contains forbidden items given that his bag triggered the alarm. First, find the number of bags did not trigger the alarm. Since Izabella's bag did not trigger the alarm, her bag is one of those $4471.$ Out of these bags, $7$ had forbidden items. Therefore, there is about a $0.16\%$ chance that Izabella's bag contains forbidden items given that her bag did not trigger the alarm.

There is about a $64.84\%$ chance that Mark's bag contains a forbidden item.

This probability is not close enough to $100\%$ to ensure that Mark's bag contains a forbidden item. Therefore, it is doubtful — but possible — that Mark's bag contains a forbidden item. Next, recall the answer found in Part D.

There is about a $0.16\%$ chance that Izabella's bag contains a forbidden item.

Since this probability is very small — less than $1\%$ — it is almost certain that Izabella does not have forbidden items in her bag — but still possible.