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7. Probability
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Integrated Math 3

7.05 Conditional probability

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Concept summary

In probability, an event is a set of outcomes of an experiment to which a probability is assigned. Two events in the same experiment can be classified as either independent or dependent events.

Independent Events

Two events are independent if the occurrence of one event does not affect the likelihood of the occurrence of the other event.

Example:

The outcomes of flipping two coins

Dependent Events

Two events are dependent if the occurrence of one event affects the likelihood of the occurrence of the other event.

Example:

The outcomes of drawing two cards in a standard deck without replacement

Conditional probability

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

Example:

The probability of drawing a diamond from a standard deck of cards given a red card has been drawn

Probability of Independent Events:

If two events, A and B, are independent, then the probability of both events occurring is the product of the probability of A and the probability of B:

\displaystyle P\left(A \cap B \right)=P\left(A\right) \cdot P\left(B\right)
\bm{P(A)}
Probability of the first event
\bm{P(B)}
Probability of the second event

Probability of Dependent Events:

For dependent events, the probability of B occurring depends on whether or not A occurred. The probability of both events occurring is the product of the probability of A and the probability of B after A occurs:

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

Conditional probability:

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). This reflects the definition of independent events, where the outcome of one event does not affect the likelihood of the occurrence of the other event.

Worked examples

Example 1

State whether the following events are independent or dependent:

a

A coin is tossed and a fair six-sided die is rolled.

Approach

To determine whether two events are independent or dependent, we should determine if the outcome of one event influences the outcome of the other event.

Solution

The outcome of tossing a coin, either heads or tails, does not affect the outcome of rolling a six-sided die. Therefore, the events are independent.

b

A card dealer randomly chooses a card from a standard deck and hides it in his pocket. The deck is then shuffled and a new card is chosen.

Approach

To determine whether two events are independent or dependent, we should determine if the outcome of one event influences the outcome of the other event.

Solution

Since the dealer hid one of the 52 cards, there are fewer cards to choose from when the second card is chosen. Therefore, the events are dependent.

Example 2

For each of the following scenarios, use probability to determine if the events are independent or dependent:

a

A color spinner has three equally sized sections labeled G, Y and R. Stella spins the spinner twice. The first spin lands on Y and the second spin lands on G.

Approach

To find the sample space of spinning the spinner twice, a tree diagram should be constructed as shown:

A tree diagram with two levels. In the first level, there are three choices: G, Y and R. Each choice from the first level is connected by lines to three choices: G, Y and R.

From the tree diagram, the sample space has 9 equally likely outcomes: \{(G, G), (G, Y), (G, R), (Y, G), (Y, Y), (Y, R), (R, G), (R, Y), (R, R) \}

To test the independence of the two events we can check if P(Y \cap G) = P(Y) \cdot P(G) is true.

Solution

\displaystyle P(Y \cap G)\displaystyle =\displaystyle \dfrac{1}{9}Only one event (Y, G) out of 9 satisfies this event
\displaystyle P(Y) \cdot P(G)\displaystyle =\displaystyle \dfrac{1}{3} \cdot \dfrac{1}{3}P(Y)=P(G)= \dfrac{1}{3} since there are 3 colors
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{1}{9}Evaluate
\displaystyle \therefore P(Y \cap G)\displaystyle =\displaystyle P(Y) \cdot P(G)Since both equal \dfrac{1}{9}

Therefore the two events are independent.

b

30 dancers audition for a part. The judges decide that 16 dancers have the right height and 20 dancers are good dancers. The events in this scenario are Right height, R, and Good dancer, G.

Approach

\\ n(R)+n(G) = 16+20 =36 \gt 30, so 6 dancers must be in R \cap G.

We can display the number of people in each category in a Venn diagram as shown:

A Venn diagram with an outer rectangle and two intersecting circles. The circle on the left is labelled as Right height and has a 10 inside it and a 6 inside the part that intersects with the circle on the right. The circle on the right is labelled as Good dancer and has a 14 inside it in addition to the 6 inside the intersection with the other circle.

To test the independence of the two events we must determine if P(R \cap G) = P(R) \cdot P(G) is true.

Solution

From the Venn diagram we see that: P(R)=\dfrac{16}{30}, \, P(G)=\dfrac{20}{30} and P(R \cap G)=\dfrac{6}{30}.

Determining if the two events are independent:

\displaystyle P(R \cap G)\displaystyle =\displaystyle \dfrac{1}{5}Simplified
\displaystyle P(R) \cdot P(G)\displaystyle =\displaystyle \dfrac{16}{30} \cdot \dfrac{20}{30}Substitute the probabilities
\displaystyle \text{}\displaystyle =\displaystyle \dfrac{16}{45}Evaluate
\displaystyle \therefore P(R \cap G)\displaystyle \neq\displaystyle P(R) \cdot P(G)

Since P(R \cap G) is not equal to the product of P(R) and P(G), the two events are dependent.

Example 3

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 substitute 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 4

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

M3.S.CP.C.5

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.*

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP2

Reason abstractly and quantitatively.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

M3.MP8

Look for and express regularity in repeated reasoning.

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