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8. Probability
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United States of AmericaTennessee
Algebra 2

8.02 Conditional probability

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Concept summary

Conditional probability

Notated as P(A \vert B), it is the probability that event A occurs given that event B has already occured.

To find the probability of event A happening given that event B already happened we can use the following formula:

\displaystyle P\left( A \vert B \right)=\dfrac{P\left( A \cap B \right)}{P\left( B \right)}
\bm{A}
The first event
\bm{B}
The second event

We read P\left( A \vert B \right) as "The probability of A given B".

We can use the conditional probability formula to determine whether two events are independent.

For two independent events, A and B, the probability of both happening is \\ P(A \cap B)=P(A)\cdot P(B).

So the conditional probability formula becomes:

\begin{aligned} P\left( \left. A \right| B \right)&=\dfrac{P(A)\cdot P(B)}{P(B)} \\ &= P(A) \end{aligned}

and:

\begin{aligned} P\left( \left. B \right| A \right)&=\dfrac{P(A)\cdot P(B)}{P(A)} \\ &= P(B) \end{aligned}

Therefore, events A and B are independent if P\left( \left. A \right| B \right)=P(A) and P\left( \left. B \right| A \right)=P(B).

For dependent events, the probability of B occuring depends on whether or not A occurred. We use the notation: P\left(A \vert B \right) to say "the probability of A given that B has occurred". The probability of both events occuring is the product of the probability of B and the probability of A after B occurs:

\displaystyle P\left(A \cap B \right)=P\left(B\right) \cdot P\left(A \vert B\right)
\bm{P(B)}
Probability of event B
\bm{P(A \vert B)}
Probability of event A given that event B has occurred

Notice that this formula is a rearrangement of the conditional probability formula.

Worked examples

Example 1

A group of people were asked whether they went on a vacation last summer. The results are provided in the given table:

VacationNo vacationTotal
Male222648
Female322052
Total5446100

Find the probability that a randomly selected person went on a vacation, given that they are male.

Approach

For this event, A represents the people that went on vacation, and B represents the people that are male.

First we should find P(\text{B}) and P(A \cap B) and then substute the values into the formula for P(A \vert B).

Solution

\displaystyle P(\text{B})\displaystyle =\displaystyle P(\text{Male})
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{48}{100}There were 48 males out of 100 people surveyed.
\displaystyle P(A \cap B)\displaystyle =\displaystyle P(\text{Vacation} \cap \text{Male})
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{22}{100}There were 22 males that went on vacation out of the 100 people surveyed.
\displaystyle P(A \vert B)\displaystyle =\displaystyle \dfrac{P\left( A \cap B \right)}{P\left( B \right)}Conditional probability formula
\displaystyle P(\text{Vacation} \vert \text{Male})\displaystyle =\displaystyle \dfrac{P\left( \text{Vacation} \cap \text{Male} \right)}{P\left( \text{Male} \right)}
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{\dfrac{22}{100}}{\dfrac{48}{100}}
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{11}{24}

The probability that a person went on a vacation, given that they are male is \dfrac{11}{24}.

Reflection

Alternatively, we could find this probability using a different method.

Because it is given that the person is a male, we are only selecting the person out of the 48 males that were surveyed. So the total number of outcomes is 48. The number of these people that went on vacation is 22. So the number of outcomes that match the event is 22. So can use the following formula:

\displaystyle P(\text{event})\displaystyle =\displaystyle \dfrac{\text{number outcomes that match the event}}{\text{total number of possible outcomes}}
\displaystyle P(\text{Vacation} \vert \text{Male})\displaystyle =\displaystyle \dfrac{22}{48}
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{11}{24}

Example 2

John selects one card from a standard deck of 52 cards:

A standard deck of 52 cards with 2 black suits, clubs and spades, and 2 red suits, hearts and diamonds. Each suit has 13 cards: Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3, and 2.

He considers the following events:

  • Event A: A black card will be selected.

  • Event B: A Jack card will be selected.

a

Describe P\left( \left. A \right| B \right).

Approach

The notation P\left( \left. A \right| B \right) means the probability of A given B. This tells us that the event B has already happened and we want to find the probability that event A happens next.

Solution

P\left( \left. A \right| B \right) is the probability that a black card was selected given that the card selected is a Jack.

b

Describe P\left( \left. B \right| A \right).

Approach

The notation P\left( \left. B \right| A \right) means the probability of B given A. This tells us that the event A has already happened and we want to find the probability that event B happens next.

Solution

P\left( \left. B \right| A \right) is the probability that a Jack was selected given that the card selected is black.

c

Describe P\left( A \cap B\right).

Approach

The notation P\left( A \cap B \right) means the probability that both A and B occur.

Solution

Given the events A and B, P\left( A \cap B\right) is the probability that both a black card and a Jack is selected. This is the probability that a black Jack is selected.

d

Determine if A and B are independent events using conditional probability.

Approach

Events A and B are independent if P\left( \left. A \right| B \right)=P(A) and P\left( \left. B \right| A \right)=P(B).

So we will need to find P(A), P(B), P(A \vert B) and P(B \vert A) to determine whether A and B are independent.

Solution

The total number of cards is 52. There are 26 black cards and 4 Jack cards. So:

\displaystyle P(A)\displaystyle =\displaystyle \dfrac{26}{52}26 black cards out of a deck of 52 cards
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{1}{2}Simplified
\displaystyle P(B)\displaystyle =\displaystyle \dfrac{4}{52}4 jacks out of a deck of 52 cards
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{1}{13}Simplified
\displaystyle P(A \cap B)\displaystyle =\displaystyle \dfrac{2}{52}2 black jacks out of a deck of 52 cards
\displaystyle P(A \vert B)\displaystyle =\displaystyle \dfrac{P(A \cap B)}{P(B)}Conditional probability formula
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{\dfrac{2}{52}}{\dfrac{4}{52}}Substitution
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{1}{2}Simplified
\displaystyle \text{}\displaystyle =\displaystyle P(A)The first condition, P\left( \left. A \right| B \right)=P(A), holds
\displaystyle P(B \vert A)\displaystyle =\displaystyle \dfrac{P(A \cap B)}{P(A)}Conditional probability formula
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{\dfrac{2}{52}}{\dfrac{26}{52}}Substitution
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{1}{13}Simplified
\displaystyle \text{}\displaystyle =\displaystyle P(B)The second condition, P\left( \left. B \right| A \right)=P(B), holds

Since P\left( \left. A \right| B \right)=P(A) and P\left( \left. B \right| A \right)=P(B), we have shown that A and B are independent events.

Outcomes

A2.S.CP.A.1

Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. Categorize events as independent or dependent.*

A2.S.CP.C.4

Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A and interpret the answer in terms of the given context.*

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP2

Reason abstractly and quantitatively.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP4

Model with mathematics.

A2.MP6

Attend to precision.

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