Conditional Probability in the Uniform Probability Model

Therefore, the following probabilities need to be calculated first. To do so, the Probability Formula will be used. Now, examine the given table to find the probabilities of each event. Adding the event labels to the table can make finding the probabilities a little easier. Recall that when finding a conditional probability, the sample space is reduced. Having found the probabilities, note that none of the conditional probabilities is equal to $P(A).$

	Fraction	Decimal
$P(A)$	$\frac{500}{1317}$	$\approx 0.380$
$P(A|B)$	$\frac{201}{324}$	$\approx 0.620$
$P(A|C)$	$\frac{118}{284}$	$\approx 0.415$
$P(A|D)$	$\frac{181}{709}$	$\approx 0.255$

Therefore, events $B,$ $C,$ and $D$ each have an effect on event $A,$ meaning that $A$ is a dependent event. In simpler terms, a passenger's chance of surviving depended on what class they were traveling in.

	$\mathbf{A}:$ Passenger Survived
$\mathbf{B}:$ First Class Passenger	Dependent, $P(A|B)\ne P(A)$
$\mathbf{C}:$ Second Class Passenger	Dependent, $P(A|C)\ne P(A)$
$\mathbf{D}:$ Third Class Passenger	Dependent, $P(A|D)\ne P(A)$

Comparing these probabilities, it can be concluded that not all passengers aboard the Titanic had the same chance of surviving. The first class passengers had the greatest chance of being rescued.

Since there are three possible outcomes, start by drawing three branches and labeling them. Additionally, since the drugs were distributed among participants evenly, each branch has the same probability, $\frac{1}{3}.$

Each of these outcomes will have two further outcomes — recovered $R$ or not recovered $NR.$ Therefore, two more branches will be drawn for each case.

Each branch should have a probability value on it. Notice that these probabilities are conditional probabilities.

Since there are only two possible outcomes after a certain medicine is received, these outcomes are complements of each other. Therefore, by the Complement Rule, the following equations can be written. Next, these conditional probabilities can be calculated using the information from the second point. Knowing that $\text{one}$ $\text{third}$ of the participants received Drug A and $7\%,$ or $0.07,$ of the participants who received Drug A recovered, $P(R|A)$ can be calculated using the Conditional Probability Formula. When a participant receives Drug A, $0.21$ of them recover, which means that $0.79$ do not recover. The other conditional probabilities can be found in a similar fashion. The percentages will be written as decimals when calculating each probability. The other conditional probabilities can be found in the table.

	Conditional Probability Formula	Substitute	Evaluate	Probability of the Complement
$P(R|A)$	$\frac{P(R\text{ and }A)}{P(A)}$	$\frac{0.07}{\frac{1}{3}}$	$0.21$	$1-0.21=0.79$
$P(R|B)$	$\frac{P(R\text{ and }B)}{P(B)}$	$\frac{0.11}{\frac{1}{3}}$	$0.33$	$1-0.33=0.67$
$P(R|C)$	$\frac{P(R\text{ and }C)}{P(C)}$	$\frac{0.05}{\frac{1}{3}}$	$0.15$	$1-0.15=0.85$

Finally, the tree diagram can be completed.

Recall that each final branch of the tree diagram represents a conditional probability. The branch between $B$ and $NR$ will represent the probability of a participant receiving Drug B and not recovering.

Therefore, the conditional probability that a participant did not recover if they received Drug B is $0.67.$

The highlighted path represents $P(A\text{ and }NR),$ the probability that a participant received Drug A and did not recover. This probability is equal to the product of the probabilities on the branches. The number of study participants who received Drug A and did not recover can be found by multiplying this probability by the total number of participants, $6000$ people. $1580$ participants received Drug A and did not recover from their condition.

Therefore, $P(S),$ $P(C|S),$ and $P(C|NS)$ are given. Notice that $NS$ is the complement of $S$ and $NC$ is the complement of $C.$ Recall that the probability of the complement of an event is $1$ minus the probability of the event. Therefore, the probabilities of the complements of the given events can be found.

With the events and probabilities determined, a tree diagram can be drawn as shown.

To find it, the tree diagram drawn in the previous part will be used. Consider all the paths that lead to event $C,$ the event that an email contains the word free. The probability of either of these two paths occurring is the sum of the products of the probabilities on the paths. Of these two paths, the topmost one leads to the favorable event. The probability of this event is the product of the probabilities on the path. As a result, the ratio of the probability of the favorable path to the probability of either of the possible paths gives $P(S|C).$ An email that contains the word free has a $0.94$ probability that it is a spam email.