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# Multiplication Rule

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Probability of Intersections

10% of the emails that Michelle receives are spam emails. Her spam filter catches spam 95% of the time. Her spam filter misidentifies non-spam as spam 2% of the time. What is the probability of an email chosen at random being spam and being correctly identified as spam by her spam filter ?

#### Guidance

Consider two events  $A$ and $B$ . The shaded section of the Venn diagram below is the outcomes shared by events  $A$ and $B$ . It is called the intersection of events  $A$ and $B$ , $A \cap B$ . Note that $B \cap A$  is equivalent to $A \cap B$ .

Sometimes you can calculate $P(A \cap B)$  directly, especially if you know all of the outcomes in the sample space. Other times, you will only have partial information about the sample space and the events.

If two events  $A$ and  $B$ are dependent , then you can find the conditional probability of  $A$ given  $B$ (or the conditional probability of  $B$ given $A$ ):

$P(A|B)=\frac{P(A \cap B)}{P(B)} \qquad P(B|A)=\frac{P(A \cap B)}{P(A)}$

You can rewrite the above equations to solve for $P(A \cap B)$ :

$P(A \cap B)=P(A|B)P(B) \quad P(A \cap B)=P(B|A)P(A)$

These formulas are known as the Multiplication Rule.

What if the events are independent ? The Multiplication Rule will still work, but it can be simplified. Recall that if two events are independent , then the result of one event has no impact on the result of the other event. For independent events  $A$ and $B$ , $P(A|B)=P(A)$  and  $P(B|A)=P(B)$ . By substitution, the above formulas become:

FOR INDEPENDENT EVENTS : $P(A \cap B)=P(A) P(B) \quad P(A \cap B)=P(B)P(A)$

Note that now, these two formulas are identical.

If you don't know whether or not two events are independent or dependent, you can always use the Multiplication Rule for calculating the probability of the intersection of the two events.  $P(A \cap B)= P(A)P(B)$ is just a special case of the Multiplication Rule.

 If the events are... You can use the formula... Independent $P(A \cap B)=P(A)P(B)$ Independent or Dependent $P(A \cap B)=P(A|B)P(B)$ or $P(A \cap B)=P(B|A)P(A)$

Remember that often you will be able to calculate $P(A \cap B)$  directly. However, sometimes you will only be given information about the conditional probabilities of the events or the probabilities of the individual events. In those cases, the Multiplication Rule is helpful.

Example A

If Mark goes to the store, the probability that he buys ice cream is 30%. The probability that he goes to the store is 10%. What is the probability of him going to the store and buying ice cream?

Solution: The first sentence of the problem is a statement of conditional probability. You could restate it as “the probability of Mark buying ice cream given that Mark has gone to the store is 30%”. Let  $S$ be the event that Mark goes to the store. Let  $I$ be the event that Mark buys ice cream. Rewrite the statements and question in the problem in terms of  $S$ and $I$ :

$P(I|S) &= 30\%=0.30\\P(S) &= 10\%=0.10\\P(S \cap I) &= ?$

In order to figure out the probability of the intersection of the events, use the Multiplication Rule.

$P(S \cap I)=P(I \cap S) &= P(I|S)P(S)\\&= 0.30 \cdot 0.10\\&= .03=3\%$

There is a 3% chance that Mark will go to the store and buy ice cream.

Example B

Consider the experiment of choosing a card from a deck, keeping it, and then choosing a second card from the deck. Let  $A$ be the event that the first card is a diamond. Let  $B$ be the event that the second card is a red card. Find $P(B \cap A)$ .

Solution: By the Multiplication Rule, $P(B \cap A)=P(B|A)P(A)$ . Consider each of these probabilities separately.

• $P(A)$  is the probability that the first card is a diamond. There are 13 diamonds in the deck of 52 cards, so $P(A)=\frac{13}{52}$ .
• $P(B|A)$  is the probability that the second card is a red card given that the first card was a diamond. After the first card was chosen, there are 51 cards left in the deck. 25 of them are red since the first card was a diamond. Therefore, $P(B|A)=\frac{25}{51}$ .

$P(B \cap A) &= P(B|A)P(A)\\&= \frac{25}{51} \cdot \frac{13}{52}\\&= \frac{25}{204} \approx 12\%$

Example C

The chance of heavy snow tomorrow is 50%, the chance of strong winds is 40%, and the chance of heavy snow or strong winds is 60%. What is the chance of a blizzard, which is heavy snow and strong winds?

Solution: This question asks for $P(heavy \ snow \cap strong \ winds)$ ; however, you were not given any conditional probabilities or indication that the events are independent. Remember that you can sometimes use the Addition Rule to solve for intersection probabilities.

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

You can rewrite this rule to solve for $P(A \cap B)$ .

$P(A \cap B)=P(A)+P(B)-P(A \cup B)$

For this problem:

$P(heavy \ snow \cap strong \ winds) &= P(heavy \ snow)+P(strong \ winds)-P(heavy \ snow \cup strong\ winds)\\$

$P(heavy \ snow \cap strong \ winds)&= 0.50+0.40-0.60\\&= 0.30$

The chance of a blizzard is 30%.

Concept Problem Revisited

10% of the emails that Michelle receives are spam emails. Her spam filter catches spam 95% of the time. Her spam filter misidentifies non-spam as spam 2% of the time. What is the probability of an email chosen at random being spam and being correctly identified as spam by her spam filter ?

To start, define two events. Let  $A$ be the event that an email is spam and let  $B$ be the event that the spam filter identifies an email as spam. Rewrite the statements and question in the problem in terms of  $A$ and $B$ .

• $P(A)=10\%=0.10$
• $P(B|A)=95\%=0.95$
• $P(B|A^\prime)=2\%=0.02$
• $P(A \cap B)=?$

By the Multiplication Rule, $P(A \cap B)=P(B|A)P(A)$ . You have enough information to use this rule to answer the question.

$P(A \cap B) &= P(B|A)P(A)\\&= 0.95 \cdot 0.10\\&= .095\\&= 9.5\%$

If an email is chosen at random, there is a 9.5% chance that it is spam and was identified as spam.

#### Vocabulary

The union of two events is the event that includes all outcomes that are in either or both of the original events. The symbol for union is $\cup$ .

The intersection of two events is the event that includes all outcomes that are in both of the original events . The symbol for intersection is $\cap$ .

The probability of an event is the chance of the event occurring.

The Addition Rule states that for two events  $A$ and $B$ , $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ .

The Multiplication Rule states that for two events  $A$ and $B$ , $P(A \cap B)=P(B|A)P(A)=P(A|B)P(B)$ .

A Venn diagram is a way to visualize sample spaces, events, and outcomes.

#### Guided Practice

1. 0.1% of the population is said to have a new disease. A test is developed to test for the disease. 98% of people without the disease will receive a negative test result. 99.5% of people with the disease will receive a positive test result. A random person who was tested for the disease is chosen. What is the probability that the chosen person does not have the disease and got a negative test result?

2. $P(C|D)=0.8$  and $P(D)=0.5$ . What is $P(C \cap D)$

3. Using the information from the previous problem, if $P(D|C)=0.6$  what is $P(C)$ ?

1. Let  $A$ be the event that a random person has the disease. Let  $B$ be the event that a random person gets a positive test result. Rewrite the statements and question in the problem in terms of $A$ and $B$ .

• $P(A)=0.1\%=0.001$
• $P(B^\prime|A^\prime)=98\%=0.98$
• $P(B|A)=99.5\%=0.995$
• $P(A^\prime \cap B^\prime)=?$

By the Multiplication Rule, $P(A^\prime \cap B^\prime)=P(B^\prime|A^\prime)P(A^\prime)$ . You know  $P(B^\prime|A^\prime)=0.98$ but you were not given $P(A^\prime)$  directly.  $P(A^\prime)+P(A)=100\%$ because these two events are complements. Therefore, $P(A^\prime)=99.9\%=0.999$ .

$P(A^\prime \cap B^\prime) &= P(B^\prime|A^\prime) P(A^\prime)\\&= 0.98 \cdot 0.999\\&= .97902\\&= 97.902\%$

This means that the probability that a random person who had this test done doesn't have the disease and got a negative test result is 97.902%. Most of the people who took the test for the disease will not have it and will get a negative test result.

2.

$P(C \cap D) &= P(C|D)P(D)\\&= 0.8 \cdot 0.5\\&= 0.4$

3. Remember that $P(C \cap D)=P(C|D)P(D)$  AND $P(C \cap D)=P(D|C)P(C)$ . Here, use the second formula and your answer to #2.

$P(C \cap D) &= P(D|C)P(C)\\0.4 &= 0.6 \cdot P(C)\\P(C) & \approx 0.67$

#### Practice

1. What is the Multiplication Rule and when is it used?

2. If events  $A$ and  $B$ are disjoint, what is $P(A \cap B)$ ?

3. If events  $A$ and  $B$ are independent, what is $P(A \cap B)$ ?

4. Why does the Multiplication Rule work for events whether or not they are dependent?

5.  $P(C)=0.4$ and $P(B|C)=0.2$ . What is $P(C \cap B)$ ?

6.  $P(A)=0.5, \ P(B)=0.7, \ P(B|A)=0.6$ . What is $P(A|B)$ ? Hint: first find $P(A \cap B)$ .

7.  $P(D)=0.3, \ P(E)=0.9, \ P(E|D)=0.7$ What is $P(D|E)$ ?

0.05% of the population is said to have a new disease. A test is developed to test for the disease. 97% of people without the disease will receive a negative test result. 99% of people with the disease will receive a positive test result. A random person who was tested for the disease is chosen.

8. What is the probability that the chosen person does not have the disease?

9. What is the probability that the chosen person does not have the disease and received a negative test result?

10. What is the probability that the chosen person does have the disease and received a negative test result?

11. If 1,000,000 people were given the test, how many of them would you expect to have the disease but receive a negative test result?

12. If Kaitlyn goes to the store, the probability that she buys blueberries is 90%. The probability of her going to the store is 30%. What is the probability of her going to the store and buying blueberries?

13. For three events $A,B$ , and $C$ , show that $P(A \cap B \cap C)=P(C|A \cap B)P(A \cap B)$ .

14. Using your answer to #14, show that $P(A \cap B \cap C)=P(C|A \cap B)P(B|A)P(A)$

15. On rainy weekend days, the probability that Karen bakes bread is 90%. On the weekend, the probability of rain is 50%. There is a 29% chance that today is a weekend day. What is the probability that today is a rainy weekend day in which Karen is baking bread?