3.5: Conditional Probabilities

What do you think is the probability that a man is over six feet tall? If you know that both of his parents were tall, would you change your estimate of the probability? If you know that both of his parents were short, would that affect your estimate in a different way? Most likely. The chance a man is over six feet tall is probably higher if he has tall parents and lower if he has short parents. A conditional probability is a probability that is based on some prior knowledge.

While conditional probability may seem like a difficult concept, we use it all the time in our every day life.

• Weather forecasters use conditional probability to predict the likelihood of future weather conditions given current conditions. They may calculate the probability of rain if it is cloudy outside.
• Sports betting companies may use conditional probability to set the odds that particular teams win their game. These odds may rely on knowledge about the team such knowing that a key player is injured.
• A doctor may use conditional probability when discussing the efficacy of a vaccine with their patient. In other words, the chance of contracting a virus may be much less when the patient takes the vaccine compared to when they do not take the vaccine.
• An insurance company uses conditional probability when setting rates for car insurance. For example, the insurance company may believe the chance you have an accident is higher if you are younger than 27.

All these examples of conditional probability have one thing in common: we assume that something is known before calculating a probability.

In Section 3.4 we learned that when events are dependent, whether or not event \(A\) occurs affects the probability that event \(B\) occurs. When using the rule \(P(A \text{ and } B)=P(A) \cdot P(B)\) with dependent events, we assume that event \(A\) has already occurred and affected the sample space of event \(B\). In the sense of probability that means that the sample space of an experiment has been restricted before finding the probability of event \(B\). We can express this idea as "the probability of \(B\), given \(A\)."

We will discuss this idea of restricting the sample space as we proceed through examples of conditional probability.

Conditional Probability

A conditional probability is the probability that an event will occur if some other condition has already occurred. This is denoted by \(P(B | A)\), which is read “the probability of \(B\) given \(A\).”

Example 1

You spin a spinner with the 8 equally likely outcomes shown below. A friend covers up the number where the spinner lands.

1. What's the probability it landed on an even number?
2. Your friend now tells you the spinner landed on a one-digit number. What's the probability it landed on an even number?
3. Your friend tells you the spinner landed on a two-digit number. What's the probability it landed on a number less than 12?

Solution

1. The sample space of spinning the spinner is \(S=\{2, 5, 6, 7, 8, 10, 12, 15\}\). Of these outcomes, 5 of them are even numbers so \(P(\text{even number})= \frac{5}{8}\).
2. Now, you have some additional knowledge about there the spinner landed. The sample space is restricted to only one-digit numbers: \(\{2, 5, 6, 7, 8\}.\) Of these outcomes, 3 of them are even numbers. The probability that the spinner landed on an even number given that it landed on a one-digit number is \(\frac{3}{5}\). We can write this as \(P(\text{even | single digit}) = \frac{3}{5}\).
3. Here, you have some different additional knowledge which restricts the original sample space to those outcomes with two digits: \(\{10, 12, 15\}.\) Of these outcomes, 1 of them is less than 12. So, \(P(\text{less than 12 | two digits}) = \frac{1}{3}\).

Example 2

One card is drawn from a well-shuffled deck of 52 cards. Find the following probabilities:

1. probability that the card is a heart given that it is red.
2. probability that the card is red given that it is a heart.
3. \(P(\text{King | face card})\)

Solution

1. We are told that the card is a red card. The sample space is restricted to only the 26 cards that are red. Of these 26 red cards, 13 are hearts. So, \(P(\text{heart | red}) = \frac{13}{26} = \frac{1}{2}\).
2. We are told that the card is a heart. The sample space is restricted to only the 13 cards that are hearts. Since every heart is red, \((P(\text{red | heart}) = \frac{13}{13} = 1\).
3. In words, \(P(\text{King | face card})\) means "probability of selecting a King given that the card is a face card." The sample space is restricted to the 12 face cards. Of these 12 face cards, 4 are Kings. \(P(\text{King | face card}) = \frac{4}{12} = \frac{1}{3}\).

Try it Now 1

A box contains a collection of ping pong balls, all the same size but of different colors and numbers as shown below.

Find these probabilities.

1. the probability of selecting a yellow ball given the ball shows a "5."
2. the probability of selecting a ball showing a number greater than 10 given the ball is black.
3. \(P(\text{odd number | not gray})\)

Answer

1. \(\frac{2}{5}\)
2. \(\frac{1}{2}\)
3. \(\frac{5}{7}\)

While we can find conditional probabilities by analyzing and restricting the sample space, it is useful to have a formula for conditional probability. Let's examine a Venn diagram to develop a general formula.

The Venn diagram shows event \(A\) and event \(B\) which overlap in the intersection of \(A \cap B\).

Suppose we want to to find the probability of event \(B\) given event \(A\), or \(P(B | A)\). This means we need to restrict the sample space to only those outcomes in event \(A\). A portion of the Venn diagram has been blackened to show that these outcomes are no longer part of the sample space, leaving only the outcomes in the given event \(A\).

The only portion of event \(B\) that remains once the sample space has been restricted to event \(A\) is the portion of the Venn diagram \(A \cap B \). Therefore, \(P(B | A) = \dfrac{n(A \cap B)}{n(A)}\) or \(\dfrac{n(A \text{ and } B)}{n(A)}\).

If we divide numerator and denominator of this conditional probability formula by the number of outcomes in the sample space \(n(S)\), then we can compute the conditional probability in alternative way as

\(P(B|A) = \dfrac{\left(\frac{n(A \text{ and } B)}{n(S)}\right)}{\left(\frac{n(A)}{n(S)}\right)} = \dfrac{P( A \text{ and } B)}{P (A)}\).

Conditional Probability Formula

For events \(A\) and \(B\), \( P(B|A) = \dfrac{n(A \text{ and } B)}{n(A)} \) or \( P(B|A) =\dfrac{P( A \text{ and } B)}{P (A)}\)

Example 3

Two fair dice are rolled and the sum of the numbers is observed. What is the probability that the sum is at least 9 if it is known that a 5 was rolled?

Solution

We are given that the dice show a 5 so this is a conditional probability. We are asked to find \(P(\text{sum is at least 9 }| \text{ 5 was rolled})\). Here, event \(A\) is "5 was rolled." Event \(B\) is "sum is at least 9."

Find the number of outcomes in event \(A\). The pairs of dice showing a 5 are

\(A = \{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6)\}.\)

Therefore, \(n(A)=11\).

List the pairs of dice in event \((A \text{ and } B)\). The pairs of dice showing 5 was rolled and the sum is at least 9 is

\(A \text{ and } B = \{(4,5), (5,5), (6,5), (5,4), (5,6)\}.\)

Therefore, \(n(A \text{ and } B)=5\).

Applying the conditional probability formula,

\( P(B|A) = \dfrac{n(A \text{ and } B)}{n(A)} = \dfrac{5}{11}\).

The probability that the sum is at least 9 if it is known that a 5 was rolled is \(\dfrac{5}{11}\).

 

Try it Now 2

A coin is flipped twice and the results are recorded. Find each probability.

1. the probability that both coins land heads given at least one coin lands heads.
2. the probability that both coins land heads given the first coin lands heads.

Answer

1. \(\frac{1}{3}\)
2. \(\frac{1}{2}\)

Example 4

For a group of people, the probability of having blond hair is 25%. The probability of having blond hair and blue eyes is 10%. What is the probability that person has blue eyes given they have blond hair?

Solution

We need to find a conditional probability because it is given that a person has blond hair. We want to find \(P(\text{ blue eyes } | \text{ blond hair })\). We don't have the sample space to count numbers of outcomes so we must use the second formula for conditional probability: \( P(B|A) =\frac{P( A \text{ and } B)}{P (A)}\).

The given event \(A\) is "blond hair," so \(B\) is the event "blue eyes." The problem states that \( P(A) = P(\text{blond hair }) = 0.25\) and \(P(A \text{ and } B) =P(\text{blond hair and blue eyes })=0.10\), so

\( P(B|A) =\dfrac{P( A \text{ and } B)}{P (A)} = \dfrac{0.10}{0.25}= \frac{2}{5} = 0.40.\)

The probability that a person has blond hair given they have blue eyes is 0.40, or 40%.

Try it Now 3

The label on a medicine bottle claims that there is a 14% chance of experiencing insomnia. There is a 5% chance of experiencing both a headache and insomnia. What's the probability that a person who takes this medicine has a headache given they have insomnia?

Answer

\(\frac{5}{14} \approx 0.3571\) or 35.71%

Example 5

The table shows survey results from 250 people who recently purchased a car.

Satisfaction of Car Buyers

	Satisfied	Not Satisfied	Total
New Car	92	28	120
Used Car	83	47	130
Total	175	75	250

Use the results in the table to find

1. the probability that a person is satisfied with their car.
2. the probability that a person is satisfied given the person bought a used car.
3. the probability that a person is not satisfied given the person bought a new car.
4. the probability that a person bought a new car given that they are satisfied.

Solution

1. The total number of customers was 250 of which 175 are satisfied with their purchase. This is not a conditional probability because the sample space has not been restricted.

So, \(P(\text{satisfied})=\frac{175}{250} = \frac{7}{10} = 0.7\)

The probability that a person is satisfied is 0.7 or 70%.

2. We are given that the customer bought a used car. This is a conditional probability. Let \(A\) represent the given condition "customer bought a used car." Let \(B\) be "customer is satisfied." We want to find \(P(B | A)\).

• The number of people who bought a used car is \(n(A)= 130\).
• The number of people who bought a used car and were satisfied is \(n(A \text{ and } B) = 83\).

So, \( P(B|A) = \frac{n(A \text{ and } B)}{n(A)} = \frac{83}{130} \approx 0.6385\).

The probability that a person is satisfied given the person bought a used car is approximately 0.6385 or 63.85%.

3. We are given that the customer bought a new car. This is a conditional probability. Let \(A\) represent the given condition "customer bought a new car." Let \(B\) be "customer is not satisfied." We want to find \(P(B | A)\).

• The number of people who bought a new car is \(n(A)= 120\).
• The number of people who bought a new car and were not satisfied is \(n(A \text{ and } B) = 28\).

So, \( P(B|A) = \frac{n(A \text{ and } B)}{n(A)} = \frac{28}{120} = \frac{7}{30} \approx 0.2333\).

The probability that a person is satisfied given the person bought a used car is approximately 0.2333 or 23.33%.

4. We are given that the customer is satisfied. This is a conditional probability. Let \(A\) represent the given condition "customer is satsified." Let \(B\) be "customer bought a new car." We want to find \(P(B | A)\).

• The number of people who are satisified is \(n(A)= 175\).
• The number of people who are satisfied and bought a new car is \(n(A \text{ and } B) = 92\).

So, \( P(B|A) = \frac{n(A \text{ and } B)}{n(A)} = \frac{92}{175} \approx 0.5257\).

The probability that a person is satisfied given the person bought a used car is approximately 0.5257 or 52.57%.

You may also see conditional probability scenarios written in ways that do not use the word "given." For example, in the previous example we could describe the conditional event "a person is satisfied given they bought a used car" as "a person is satisfied if they bought a used car" or "a person who bought a used car is satisfied."

Example 6

A survey of 350 students at a university revealed the following data about class standing and place of residence.

Living Arrangements of University Students

	Freshman	Sophomore	Junior	Senior	Total
Dormitory	89	34	46	15	184
Apartment	32	17	22	48	119
with Parents	13	31	3	0	47
Total	134	82	71	63	350

Use the results in the table to find each probability.

1. What is the probability that a student is a sophomore if the student lives in an apartment?
2. What is the probability that a student lives with parents if the student is a freshman?

Solution

1. We are given that the student lives in an apartment. This is a conditional probability. Let \(A\) represent the given condition "student lives in an apartment." Let \(B\) be "student is a sophomore." We want to find \(P(B | A)\).

• The number of students who live in an apartment is \(n(A)= 119\).
• The number of students who live in an apartment and are sophomores is \(n(A \text{ and } B) = 17\).

So, \( P(B|A) = \frac{n(A \text{ and } B)}{n(A)} = \frac{17}{119} \approx 0.1429\).

The probability that a student is a sophomore if they live in an apartment is approximately 0.1429 or 14.29%.

2. We are given that the student is a freshman. This is a conditional probability. Let \(A\) represent the given condition "student is a freshman." Let \(B\) be "student lives with parents." We want to find \(P(B | A)\).

• The number of students who are freshmen is \(n(A)= 134\).
• The number of people who are freshmen and live with parents is \(n(A \text{ and } B) = 13\).

So, \( P(B|A) = \frac{n(A \text{ and } B)}{n(A)} = \frac{13}{134} \approx 0.0970\).

The probability that a lives with parents if they are a freshman is approximately 0.0970 or 9.70%.

Try it Now 4

A group of people were surveyed about the type of movies they prefer. Suppose a person is chosen at random from this group.

Movie Preferences

Gender	Romantic, \(R\)	Action, \(A\)	Horror, \(H\)	Total
Male, \(M\)	8	25	6	39
Female, \(F\)	12	10	3	25
Total	20	35	9	64

Use the results in the table to find

1. the probability that a person prefers action movies given they are male.
2. the probability that a person is female given they prefer romantic movies.

Answer

1. \(\frac{25}{39}\)
2. \(\frac{12}{20}= \frac{3}{5}\)