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# Conditional Probability and Independence

## Explore formulas to determine probability of combined events

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Independent, Conditional, and Mutually Exclusive Events

Freezy's Ice Cream Stand doesn't think it has enough information to decide if it should add Pumpernickel Brickel or Dandy Cotton Candy to its menu. Therefore, it conducts another poll of its customers. It finds that the probability that a customer will like both flavors is 0.33 and the probability that a customer will like the Cotton Candy flavor is 0.8. What is the probability a customer will like the Pumpernickel flavor?

### Independent Conditional and Mutually Exclusive Events

We have already discussed the formula for conditional probabilities: \begin{align*}P(A|B)=\frac{P(A \cap B)}{P(B)}\end{align*}. These events are not independent, they are conditional because the outcome of event \begin{align*}B\end{align*} affects the outcome of event \begin{align*}A\end{align*}. When events are independent, \begin{align*}P(A|B)=P(A)\end{align*}, meaning that it doesn’t matter that event \begin{align*}B\end{align*} has occurred, the result of event \begin{align*}B\end{align*} does not affect the result of event \begin{align*}A\end{align*}. Now we can replace \begin{align*}P(A|B)\end{align*} with \begin{align*}\frac{P(A \cap B)}{P(B)}\end{align*} in the previous statement to get \begin{align*}\frac{P(A \cap B)}{P(B)}=P(A)\end{align*}. Finally, multiply both sides by \begin{align*}P(B)\end{align*} to get \begin{align*}P(A \cap B)=P(A) \times P(B)\end{align*} for independent events \begin{align*}A\end{align*} and \begin{align*}B\end{align*}. We can use this rule to determine if events are independent or to find the intersection of known independent events.

It is also possible for two events to have not intersection, or \begin{align*}P(A \cap B)=0\end{align*}. When this occurs we say that the events (or sets) are Mutually Exclusive. If one has occurred, then the other cannot occur. Examples of Mutually Exclusive sets are Boys and Girls, Juniors and Seniors-it is not possible to be both. It is important to note here that mutually exclusive events cannot be independent unless the probability of one of the events is zero since for independent events \begin{align*}P(A \cap B)=P(A) \times P(B)\end{align*} and the only way a product can equal zero is if one of the factors is equal to zero.

#### Solve the following problems

Given two events, \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, such that \begin{align*}P(A)=0.3\end{align*}, \begin{align*}P(B)=0.5\end{align*} and \begin{align*}P(A \cup B)=0.65\end{align*}

Find \begin{align*}P(A \cap B)\end{align*}.

Since we are not told that the events are independent, we do not know that \begin{align*}P(A \cap B)=P(A) \times P(B)\end{align*}. However, for all events, independent or otherwise, it is true that \begin{align*}P(A)+P(B)-P(A \cap B)=P(A \cup B)\end{align*} so

\begin{align*}0.3+0.5-P(A \cap B)&=0.65 \\ P(A \cap B)&=0.15\end{align*}

State with a reason whether the events are independent.

To determine if the events are independent we will test the rule \begin{align*}P(A \cap B)=P(A) \times P(B)\end{align*}.

\begin{align*}P(A) \times P(B)=0.3 \times 0.5=0.15=P(A \cap B).\end{align*}

Thus, the events are independent since the product of their probabilities is equal to the probability of their intersection.

State with a reason whether the events are mutually exclusive.

The events are not mutually exclusive because \begin{align*}P(A \cap B)=0.15 \ne 0\end{align*}.

#### Given that \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent events such that \begin{align*}P(A)=0.4\end{align*} and \begin{align*}P(A \cup B)=0.76\end{align*}, find

\begin{align*}P(B)\end{align*}

Since we know the two events are independent, we know that \begin{align*}P(A \cap B)=0.4 P(B)\end{align*}. Now we can use the formula for the probability of the union of the two sets and substitute this product for the probability of the intersection:

\begin{align*}0.4+P(B)-0.4 P(B)&=0.76 \\ 0.6 P(B)&=0.36 \\ P(B)&=0.6\end{align*}

Probability of \begin{align*}A\end{align*} or \begin{align*}B\end{align*} but not both occurring.

To find the probability of either \begin{align*}A\end{align*} or \begin{align*}B\end{align*} occurring but not both, we need to find \begin{align*}P(A \cap B)\end{align*} and subtract this from \begin{align*}P(A \cup B)\end{align*}.

\begin{align*}0.76-(0.4)(0.6)=0.76-0.24=0.52\end{align*}

#### Events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent such that \begin{align*}P(B \cap A^\prime)=0.2\end{align*} and \begin{align*}P(A \cap B)=0.3\end{align*}. Find \begin{align*}P(A \cup B)\end{align*}.

For this problem, a Venn diagram might be useful to illustrate the given information.

From the diagram we can see \begin{align*}P(B)=0.5\end{align*} and since we know that the events are independent, we know:

\begin{align*}P(A) \times P(B)&=P(A \cap B) \\ P(A) \times 0.5&=0.3 \\ P(A)&=\frac{0.3}{0.5}=0.6\end{align*}

Now, \begin{align*}P(A \cup B)=0.6+0.5-0.3=0.8\end{align*}.

### Examples

#### Example 1

Earlier, you were asked what is the probability a customer will like the Pumpernickel flavor.

Since we know liking the Pumpernickel flavor (B) and liking the Cotton Candy flavor (A) are independent events, we also know that \begin{align*}P(A \cap B)=0.33 P(B)\end{align*}. Now we can use this formula for the probability of the intersection of the two sets to find the information we're seeking.

\begin{align*}0.33=0.8 P(B) \\ P(B)&=0.4125\end{align*}

Therefore, the probability that a customer will like the Cotton Candy flavor is 41.25%.

#### Example 2

Given two events, \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, such that \begin{align*}P(A)=0.4\end{align*}, \begin{align*}P(B)=0.5\end{align*} and \begin{align*}P(A \cup B)=0.75\end{align*}

Find \begin{align*}P(A \cap B)\end{align*}.

\begin{align*}0.4+0.5-P(A \cap B)&=0.75 \\ P(A \cap B)&=0.15\end{align*}

State with a reason whether the events are independent.

If the events are independent, \begin{align*}P(A \cap B)=0.4 \times 0.5=0.2\end{align*}.

Since \begin{align*}0.2 \ne 0.15\end{align*}, the events are not independent.

Find \begin{align*}P(A|B)\end{align*}.

\begin{align*}P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{0.15}{0.5}=0.3\end{align*}.

#### Example 3

Given that \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent events such that \begin{align*}P(A)=0.8\end{align*} and \begin{align*}P(A \cup B)=0.9\end{align*}, find

\begin{align*}P(B)\end{align*}

\begin{align*}0.8+P(B)-0.8 P(B)&=0.9 \\ 0.2 P(B)&=0.1 \\ P(B)&=0.5\end{align*}

\begin{align*}P(B \cap A^\prime)\end{align*}

\begin{align*}P(B \cap A^\prime)=P(B)-P(B \cap A)=0.5-0.8 \times 0.5=0.1\end{align*}

#### Example 4

Events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent such that \begin{align*}P(A \cap B^\prime)=0.25\end{align*} and \begin{align*}P(A \cap B)=0.25\end{align*}. Find \begin{align*}P(A \cup B)\end{align*}.

\begin{align*}P(A)=P(A \cap B^\prime)+P(A \cap B)=0.25+0.25=0.5\end{align*}

\begin{align*}P(A \cap B)=0.5 P(B)=0.25\end{align*} and thus \begin{align*}P(B)=0.5\end{align*}

\begin{align*}P(A \cup B)=P(A)+P(B)-P(A \cap B)=0.5+0.5-0.25=0.75\end{align*}

### Review

1. Events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are mutually exclusive. Describe \begin{align*}P(A|B)\end{align*}.
2. Events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent. Show that \begin{align*}P(B)=P(B|A)\end{align*}.

For problems 3-6, use the given information about events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} to determine whether or not the events are independent.

1. \begin{align*}P(A)=0.6\end{align*}, \begin{align*}P(B)=0.4\end{align*} and \begin{align*}P(A \cup B)=0.76\end{align*}
2. \begin{align*}P(A)=0.5\end{align*}, \begin{align*}P(A \cap B)=0.2\end{align*} and \begin{align*}P(A \cup B)=0.8\end{align*}
3. \begin{align*}P(A)=0.3\end{align*}, \begin{align*}P(B)=0.4\end{align*} and \begin{align*}P(A \cup B)=0.55\end{align*}
4. \begin{align*}P(A)=0.6\end{align*}, \begin{align*}P(B \cap A^\prime)=0.28\end{align*} and \begin{align*}P(A \cap B)=0.42\end{align*}

For problems 7-10, events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent.

1. Given \begin{align*}P(A)=0.8\end{align*} and \begin{align*}P(A \cup B)=0.88\end{align*}, find \begin{align*}P(B)\end{align*} and \begin{align*}P(A \ or \ B \ \text{but not both})\end{align*}.
2. Given \begin{align*}P(A \cap B^\prime)=0.54\end{align*} and \begin{align*}P(A \cap B)=0.36\end{align*}, find \begin{align*}P(B)\end{align*} and \begin{align*}P(A \cup B)\end{align*}.
3. Given \begin{align*}P(B)=0.8\end{align*} and \begin{align*}P(A^\prime \cap B^\prime)=0.04\end{align*}, find \begin{align*}P(A)\end{align*} and \begin{align*}P(A^\prime \cup B^\prime)\end{align*}.
4. \begin{align*}P(A \cap B)=0.28\end{align*} and \begin{align*}P(A \cup B)=0.82\end{align*}, find \begin{align*}P(A)\end{align*} and \begin{align*}P(B)\end{align*}.

To see the Review answers, open this PDF file and look for section 12.14.

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