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4.4: Two Basic Rules of Probability

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When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.

The Multiplication Rule

If $A$ and $B$ are two events defined on a sample space, then: $P(A \cap B)=P(B) P(A \mid B)$. We can think of the intersection symbol as substituting for the word "and".

This rule may also be written as: $P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}$

This equation is read as the probability of $A$ given $B$ equals the probability of $A$ and $B$ divided by the probability of $B$.

If $A$ and $B$ are independent, then $P(A \mid B)=P(A)$. Then $P(A \cap B)=P(A \mid B) P(B)$ becomes $P(A \cap B)=P(A)(B)$ because the $P(A \mid B)=P(A)$ if $A$ and $B$ are independent.

The Addition Rule

If $A$ and $B$ are defined on a sample space, then: $P(A \cup B)=P(A)+P(B)-P(A \cap B)$. We can think of the union symbol substituting for the word "or". The reason we subtract the intersection of $A$ and $B$ is to keep from double counting elements that are in both $A$ and $B$.

If $A$ and $B$ are mutually exclusive, then $P(A \cap B)=0$. Then $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ becomes $P(A \cup B)=P(A)+P(B)$.

Exercise $\PageIndex{1}$

Kelsey is a community health coordinator trying to choose which pilot program to implement in her district. Her two choices are: A= a youth obesity prevention program and B= a senior fall-prevention initiative.

Kelsey can only afford to fund one program this year. The probability that she chooses A is P(A)=0.6 and the probability that she chooses B is P(B)=0.35.

P(A∩B)=0 because the budget only allows for a single program (they are mutually exclusive).

Therefore, the probability that she chooses either the youth program or the senior initiative is P(A∪B)=P(A)+P(B)=0.6+0.35=0.95. Note that the probability that she does not choose to implement either program (perhaps due to further budget cuts) must be 0.05.

$P (A O R B)$

Exercise $\PageIndex{2}$

Carlos is a collegiate soccer player. Based on his season statistics, he successfully scores a goal on 65% of his shots. During the next match, Carlos is going to attempt two shots in a row. Let A = the event Carlos is successful on his first attempt (P(A)=0.65). Let B = the event Carlos is successful on his second attempt (P(B)=0.65).

In kinesiology, we often study "momentum" or "streak shooting." For Carlos, the probability that he makes the second goal GIVEN that he made the first goal increases to 0.90 (P(B∣A)=0.90).

a. What is the probability that he makes both goals?

b. What is the probability that Carlos makes either the first goal or the second goal?

c. Are A and B independent?

d. Are A and B mutually exclusive?

Answer

a. The problem is asking you to find P(A∩B). Since we know the conditional probability P(B∣A)=0.90, we use the multiplication rule:

P(B∩A)=P(B∣A)×P(A)=(0.90)(0.65)=0.585

Carlos makes both the first and second goals with a probability of 0.585.

b. The problem is asking you to find P(A∪B). We use the general addition rule to account for the overlap:

P(A∪B)=P(A)+P(B)−P(A∩B)=0.65+0.65−0.585=0.715

Carlos makes either the first goal or the second goal with a probability of 0.715.

c. No, they are not independent. To be independent, P(B∩A) must equal P(B)×P(A).

(0.65)(0.65)=0.4225

Since 0.4225 does not equal 0.585, the events are dependent. His success on the first shot significantly "boosts" his probability of success on the second.

d. No, they are not mutually exclusive because P(A∩B)=0.585. To be mutually exclusive, the probability of both occurring must be zero.

Solution

a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = 0.90: P(B AND A) = P(B|A) P(A) = (0.90)(0.65) = 0.585

Carlos makes the first and second goals with probability 0.585.

b. The problem is asking you to find P(A OR B).

P(A OR B) = P(A) + P(B) - P(A AND B) = 0.65 + 0.65 - 0.585 = 0.715

Carlos makes either the first goal or the second goal with probability 0.715.

c. No, they are not, because P(B AND A) = 0.585.

P(B)P(A) = (0.65)(0.65) = 0.423

0.423 ≠ 0.585 = P(B AND A)

So, P(B AND A) is not equal to P(B)P(A).

d. No, they are not because P(A and B) = 0.585.

To be mutually exclusive, P(A AND B) must equal zero.

Solution

a. 0.98; b. 0.1401; c. 0.857; d. 0.15

Exercise $\PageIndex{3}$

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.

a. What is the probability that the member is a novice swimmer?

b. What is the probability that the member practices four times a week?

c. What is the probability that the member is an advanced swimmer and practices four times a week?

d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?

e. Are being a novice swimmer and practicing four times a week independent events? Why or why not?

Answer

a. $\dfrac{28}{150}$

b. $\dfrac{80}{150}$

c. $\dfrac{40}{150}$

d. $P($ advanced $\cap$ intermediate $)=0$, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.

e. No, these are not independent events.
$P$ (novice $\cap$ practices four times per week) $=0.0667$
$P$ (novice) $P$ (practices four times per week) $=0.0996$
$0.0667 \neq 0.0996$

Solution

a. 2815028150" role="presentation" style="position: relative;"> $\frac{28}{150}$

b. 8015080150" role="presentation" style="position: relative;"> $\frac{80}{150}$

c. 4015040150" role="presentation" style="position: relative;"> $\frac{40}{150}$

d. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.

e. No, these are not independent events.
P(novice AND practices four times per week) = 0.0667
P(novice)P(practices four times per week) = 0.0996
0.0667 ≠ 0.0996

Solution

a. 0.98; b. 0.1401; c. 0.857; d. 0.15

Exercise $\PageIndex{5}$

Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.

What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
Given that the woman has breast cancer, what is the probability that she tests negative?
What is the probability that the woman has breast cancer AND tests negative?
What is the probability that the woman has breast cancer or tests negative?
Are having breast cancer and testing negative independent events?
Are having breast cancer and testing negative mutually exclusive?

Answer

a. $P(B)=0.143 ; P(N)=0.85$

b. $P(N \mid B)=0.02$

c. $P(B \cap N)=P(B) P(N \mid B)=(0.143)(0.02)=0.0029$

d. $P(B \cup N)=P(B)+P(N)-P(B \cap N)=0.143+0.85-0.0029=0.9901$

e. No. $P(N)=0.85 ; P(N \mid B)=0.02$. So, $P(N \mid B)$ does not equal $P(N)$.

f. No. $P(B \cap N)=0.0029$. For $B$ and $N$ to be mutually exclusive, $P(B \cap N)$ must be

Solution

a. P(B) = 0.143; P(N) = 0.85

b. P(N|B) = 0.02

c. P(B AND N) = P(B)P(N|B) = (0.143)(0.02) = 0.0029

d. P(B OR N) = P(B) + P(N) - P(B AND N) = 0.143 + 0.85 - 0.0029 = 0.9901

e. No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N).

f. No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero.

Solution

a. 0.98; b. 0.1401; c. 0.857; d. 0.15

Exercise $\PageIndex{6}$

Refer to the information in Example $\PageIndex{5}$. $P=$ tests positive.

a. Given that a woman develops breast cancer, what is the probability that she tests positive. Find $P(P \mid B)=1-P(N \mid B)$.
b. What is the probability that a woman develops breast cancer and tests positive. Find $P(B \cap P)=$ $P(P \mid B) P(B)$.
c. What is the probability that a woman does not develop breast cancer. Find $P(B)=1-P(B)$.
d. What is the probability that a woman tests positive for breast cancer. Find $P(P)=1-P(N)$.

Answer: a. 0.98; b. 0.1401; c. 0.857; d. 0.15

Solution

a. 0.98; b. 0.1401; c. 0.857; d. 0.15

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The Multiplication Rule

The Addition Rule

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$P (A O R B)$

Exercise \(\PageIndex{2}\)

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Exercise \(\PageIndex{3}\)

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Exercise \(\PageIndex{5}\)

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Exercise \(\PageIndex{6}\)

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