AP Statistics Topic 2.6: Conditional Probability

01 · Start here

The concept in 60 seconds

Conditional probability is the chance of an event within a specified group. The word “given” tells you which outcomes remain eligible.

Suppose $1$ ticket is drawn uniformly from numbers $1\text{-}8$ . You are told that its number is at least $6$ . Only $6$ , $7$ and $8$ remain possible under that information. To find the chance of an even number, count $2$ even results among those $3$ eligible results.

Ask “among which group?”
Among tickets numbered at least $6$ , $2\text{ of the }3$ numbers are even. The conditional probability is $\frac{2}{3}$ .

Given “at least

6

”: keep

3

eligible numbers

$1$ Outside

B

$2$ Outside

B

$3$ Outside

B

$4$ Outside

B

$5$ Outside

B

$6$ In

B

+ even $7$ In

B

+ odd $8$ In

B

+ even

Eligible group

B

$6$ , $7$ , $8$

6

7

8

$3$ equally likely possibilities after using the information.

Target within

B

Even: $6$ and $8$

6

8

$2\text{ of }3\text{ eligible outcomes}\to\frac23$

“Outside $B$ ” numbers do not enter this conditional denominator. Dark tickets satisfy both $B$ and the even target; text labels identify the groups as well as color.

The original bag still has $8$ tickets. The information changes the group used in the calculation; it does not physically remove tickets.

Quick check: why is the denominator

3

instead of

8

We are calculating the chance of even given that the number is at least $6$ . The $3$ eligible outcomes are $6$ , $7$ and $8$ . The denominator must describe that given group.

02 · State the events and condition

The ticket question

A bag contains $8$ identical, well-mixed tickets numbered $1\text{-}8$ , $1\text{ of each}$ . $1$ ticket is drawn without looking. All $8$ numbered tickets are equally likely.

$A$ : the number is even, so $A = \{2, 4, 6, 8\}$ .
$B$ : the number is at least $6$ , so $B = \{6, 7, 8\}$ .
$A \cap B$ : both conditions hold, so $A \cap B = \{6, 8\}$ .

Question: Given that the drawn number is at least $6$ , what is the probability that it is even?

The target is $A$ , and the given event is $B$ . We therefore want $P(A\mid B)$ . The eligible group is $B$ ; within it, the successful outcomes are the shared outcomes $A \cap B$ .

Knowing $B$ does not make all $8$ tickets eligible for this calculation, and it does not tell us that $A$ must occur: ticket $7$ is still possible.

03 · Read the notation before calculating

Key ideas and notation

Target event: $A$

The event whose chance we want.

In $P(A\mid B)$ , $A$ is on the left of the bar.

Given event: $B$

The information that restricts the eligible group.

In $P(A\mid B)$ , $B$ is on the right of the bar.

Intersection: $A \cap B$

The outcomes satisfying both the target and the condition.

Tickets $6$ and $8$ are both even and at least $6$ .

Conditional probability

The share of the given group that also satisfies the target.

$P(A\mid B) = \frac{2}{3}$ for the ticket question.

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

Provided $P(B)>0$ .

Read the bar $\mid$ as “given,” not as division. The fraction on the right shows how to calculate the probability.

For equally likely individual outcomes, the whole-space denominator cancels:

P(A\mid B)=\frac{\text{number of outcomes in }A\cap B}{\text{number of outcomes in }B}

Ticket example: $\frac28\div\frac38=\frac23\approx0.6667=66.67\%$ .

Three questions: overall, joint and conditional

Overall ·

P(A)

Even

Given group: all $8$ tickets
Target count: $4$

$\frac{4}{8} = 0.50$

Joint ·

P(A\cap B)

Even and at least $6$

Given group: all $8$ tickets
Shared count: $2$

$\frac{2}{8} = 0.25$

Conditional ·

P(A\mid B)

Even given at least $6$

Given group: $3$ tickets in $B$
Shared count: $2$

$\frac{2}{3} \approx 0.6667$

$A$ : even; $B$ : at least $6$ . Joint and conditional probabilities use the same shared outcomes, but they describe different eligible groups.

If the outcomes have unequal probabilities, use their probability weights in the formula. Counting labels alone can give the wrong answer.

04 · Use the given group’s total

Read a conditional probability from a two-way table

Use the same original $60$ -student teaching table as in Topic 2.5. A student can belong to both Chess and Music. $1\text{ of the}$ $60$ listed students is selected uniformly at random.

Let $C$ mean “Chess member” and $M$ mean “Music member.” The table shows how these two memberships overlap.

Same shared count, different given-group totals

Original $60$ -student club example. The shared cell is $12$ ; conditional denominators are the relevant group totals.
Chess member?	Music Yes	Music No	Row total
Yes	$12$	$18$	$30$
No	$9$	$21$	$30$
Total	$21$	$39$	$60$

Given Chess ·

C

Music among Chess members

Shared count: $12$
Chess total: $30$

$P(M\mid C) = \frac{12}{30} = 40\%$

Given Music ·

M

Chess among Music members

Shared count: $12$
Music total: $21$

$P(C\mid M) = \frac{12}{21} \approx 57.14\%$

$C$ : Chess member; $M$ : Music member. This is an original fictional teaching table. All probabilities refer to $1\text{ uniform selection}$ from the listed students.

Given Chess: use the Chess row

There are $30$ Chess members, and $12$ of them are also Music members. Thus $P(M\mid C) = \frac{12}{30} = 0.40$ .

Given Music: use the Music column

There are $21$ Music members, and $12$ of them are also Chess members. Thus $P(C\mid M) = \frac{12}{21} = \frac{4}{7} \approx 0.5714$ .

The shared numerator is $12$ in both calculations, but the given groups differ. That is why $P(M\mid C)$ and $P(C\mid M)$ need not be equal. By contrast, the joint probability $P(C\cap M)$ is $\frac{12}{60} = 0.20$ .

A reliable table routine: circle the word after “given,” find that row or column total, then divide the shared cell by that total.

05 · Keep the eligible group clear

Keep the condition fixed

Complements within the given group

Among Chess members, a student either belongs to Music or does not. These two possibilities cover the same $30$ -student given group.

Keep “given Chess” fixed: Music versus not Music

Eligible group: all $30$ Chess members

Olive segment · Music $12$ of $30$

$P(M\mid C) = \frac{12}{30} = 0.40$

Light segment · not Music $18$ of $30$

$P(M^c\mid C) = \frac{18}{30} = 0.60$

The bar divides the same given group into target and target complement. Its segment widths are $40\%$ and $60\%$ . The condition stays Chess in both probabilities.

$P(A^c\mid B) = 1 – P(A\mid B)$ , provided $P(B) > 0$ .

The complement changes the target on the left, while the condition $B$ stays fixed. $P(A\mid B^c)$ asks about a different given group and cannot generally be found by subtracting $P(A\mid B)$ from $1$ .

A zero-probability condition

The ratio formula requires $P(B) > 0$ . If $P(B) = 0$ , division by $P(B)$ is undefined. In the finite ticket model, “given the number is $9$ ” is not an eligible condition because no ticket $9$ exists.

A conditional probability of $0$ is possible when the given group has positive probability but contains no target outcomes. For example, on a fair die, $P(\text{odd}\mid \text{even}) = 0$ because the $3\text{ even outcomes}$ contain no odd number.

Information does not establish a cause

A conditional percentage describes a group. Saying that $40\%$ of Chess members belong to Music does not show that Chess membership caused Music membership.

06 · Connect a condition to a joint probability

Use the general multiplication rule

Rearranging the conditional probability formula gives the chance that both events occur:

$P(A\cap B) = P(A) \times P(B\mid A)$ .

Equivalently, $P(A\cap B) = P(B) \times P(A\mid B)$ .

The factor after the multiplication sign must use the first factor’s event as its condition. That conditioning event must have positive probability. In the club example, $P(C\cap M) = (\frac{30}{60}) \times (\frac{12}{30}) = \frac{12}{60} = 0.20$ .

This is the general rule. It does not require the events to be independent. Do not replace $P(B\mid A)$ with $P(B)$ unless the event relationship justifies it.

Draws without replacement: update what remains

A bag contains $3$ Green tokens and $2$ Gold tokens, identical in size and thoroughly mixed. $2$ tokens are drawn sequentially without replacement; each remaining token is equally likely at each draw. $1$ complete trial consists of both draws.

Without replacement: the first draw changes the bag

Initial bag ·

3\text{ Green}+2\text{ Gold}=5\text{ tokens}

GreenGreenGreenGoldGold

After first Green $2\text{ Green}+2\text{ Gold remain}$

GreenGreenGoldGold

Second Green given first Green: $\frac{2}{4}$
Second Gold given first Green: $\frac{2}{4}$

After first Gold $3\text{ Green}+1\text{ Gold remain}$

GreenGreenGreenGold

Second Green given first Gold: $\frac{3}{4}$
Second Gold given first Gold: $\frac{1}{4}$

Each chip represents $1$ token. Labels distinguish Green and Gold; both counts and the remaining total matter for the second-draw probabilities.

Let $G_1$ mean “first token is Green” and $G_2$ mean “second token is Green.” After a first Green token, $2$ Green tokens remain among $4$ tokens, so $P(G_2\mid G_1) = \frac{2}{4}$ .

$P(G_1\cap G_2) = (\frac{3}{5}) \times (\frac{2}{4}) = \frac{3}{10} = 0.30$ .

4

ordered paths for

2

draws without replacement

First drawGreen · probability $\frac{3}{5}$

Then · second draw given first GreenGreen · conditional probability $\frac{2}{4}$

Ordered path: $\text{Green}\to\text{Green}$
$\frac{3}{5} \times \frac{2}{4} = 0.30$

Then · second draw given first GreenGold · conditional probability $\frac{2}{4}$

Ordered path: $\text{Green}\to\text{Gold}$
$\frac{3}{5} \times \frac{2}{4} = 0.30$

First drawGold · probability $\frac{2}{5}$

Then · second draw given first GoldGreen · conditional probability $\frac{3}{4}$

Ordered path: $\text{Gold}\to\text{Green}$
$\frac{2}{5} \times \frac{3}{4} = 0.30$

Then · second draw given first GoldGold · conditional probability $\frac{1}{4}$

Ordered path: $\text{Gold}\to\text{Gold}$
$\frac{2}{5} \times \frac{1}{4} = 0.10$

Each second-draw probability is conditional on the first-draw result. Multiply along a path to find its joint probability. The $4$ complete path probabilities total $0.30 + 0.30 + 0.30 + 0.10 = 1$ .

Read a path from its first result to its second result, then multiply along that path. Second-draw probabilities are conditional on the first result.

With replacement and thorough mixing between draws, the bag would return to $3$ Green and $2$ Gold tokens. Then the Green–Green probability would be $(\frac{3}{5}) \times (\frac{3}{5}) = 0.36$ . Replacement changes the model, so name it before calculating.

07 · Follow the reasoning

Worked examples

Example 1: restrict an outcome list

Question: $1$ ticket is drawn uniformly from numbers $1\text{-}8$ . Given that its number is at least $6$ , what is the probability that it is even?

Given group $B = \{6, 7, 8\}$ .
Target outcomes within $B$ are $\{6, 8\}$ .
$P(\text{even}\mid \text{at least 6}) = \frac{2}{3} \approx 0.6667$ .

Interpretation: among the $3$ eligible numbers at least $6$ , $2\text{ are even}$ . The answer is about $66.67\%$ , rather than the joint probability $\frac{2}{8} = 25\%$ .

Example 2: reverse the condition carefully

Question: Use the $60$ -student club table to compare $P(\text{Music}\mid \text{Chess})$ and $P(\text{Chess}\mid \text{Music})$ .

Given Chess, the eligible group has $30$ students: $P(M\mid C) = \frac{12}{30} = 0.40$ .
Given Music, the eligible group has $21$ students: $P(C\mid M) = \frac{12}{21} = \frac{4}{7} \approx 0.5714$ .

Check: both use the shared count $12$ . Reversing the bar changes the denominator, so the answers differ.

Example 3: divide given probabilities

Question: A fictional course model gives $P(\text{Pass}\cap \text{Attend}) = 0.42$ and $P(\text{Attend}) = 0.70$ . Find $P(\text{Pass}\mid \text{Attend})$ .

The condition is Attend, so its probability is the denominator.
$P(\text{Pass}\mid \text{Attend}) = \frac{0.42}{0.70} = 0.60$ .

Interpretation: within the Attend event, the model’s pass probability is $60\%$ . The joint probability $42\%$ describes both events within the whole model.

Example 4: multiply using the stated condition

Question: A workshop model gives $P(\text{Register}) = 0.30$ and $P(\text{Bring laptop}\mid \text{Register}) = 0.40$ . Find the probability of registering and bringing a laptop.

The target is the joint event $\text{Register}\cap\text{Bring laptop}$ .
Use $P(\text{Register}) \times P(\text{Bring laptop}\mid \text{Register})$ .
The joint probability is $0.30 \times 0.40 = 0.12$ .

Interpretation: the model gives a $12\%$ chance of both events. The value $0.40$ is a conditional probability; it is not automatically the overall probability of bringing a laptop.

Example 5: $2\text{ Green tokens}$ without replacement

Question: $2$ tokens are drawn without replacement from the bag with $3$ Green and $2$ Gold tokens. Find the probability that both are Green.

First Green probability: $\frac{3}{5}$ .
After a first Green, the remaining bag contains $2$ Green and $2$ Gold tokens.
Second Green given first Green: $\frac{2}{4}$ .
$P(\text{both Green}) = (\frac{3}{5}) \times (\frac{2}{4}) = 0.30$ .

Check: the second numerator and denominator both change. Using $\frac{3}{5}$ again would ignore the removal of the first token.

Example 6: a conditional complement

Question: Given that the selected student belongs to Chess, what is the probability that the student does not belong to Music?

The condition remains Chess, with $30$ eligible students.
There are $18$ Chess members who are not in Music.
$P(M^c\mid C) = \frac{18}{30} = 0.60$ .

Alternative: $1 – P(M\mid C) = 1 – 0.40 = 0.60$ . Subtracting the joint probability $0.20$ from $1$ would answer a different question.

08 · Name the group in your answer

Explain in context

Start your interpretation with “Among…” or “Given that…”. That makes the denominator visible in your words as well as in your calculation.

A complete table response: “Given that the selected student is a Chess member, the eligible group contains $30$ students. $12\text{ also}$ belong to Music, so $P(\text{Music}\mid \text{Chess}) = \frac{12}{30} = 0.40$ . Among Chess members in this listed group, $40\%$ are Music members.”

A good multiplication response identifies the first event probability and the appropriate conditional probability, then interprets their product as the chance that both events occur.

Improve this answer: “The Music probability is

40\%

.”

The statement omits its condition. A clearer answer is: “The probability of Music membership given Chess membership is $40\%$ .” Overall Music membership in this table has probability $\frac{21}{60} = 35\%$ , so the two statements describe different groups.

Before finishing: check the direction of the bar, identify the given group, show the ratio or conditional product, and keep the condition in the interpretation.

09 · Find and fix mistakes

Find and fix mistakes

Common errors and the reasoning that fixes them.
Mistake	Better reasoning
Use the whole table total for every probability.	For $P(\text{Music}\mid \text{Chess})$ , use the $30$ Chess members, not all $60$ students.
Reverse $P(A\mid B)$ and $P(B\mid A)$ .	Read the right side of the bar first. It determines the denominator.
Use $\frac{P(A)}{P(B)}$ in the conditional formula.	Use the shared numerator $P(A\cap B)$ , not the target probability alone.
Treat the bar $\mid$ as a multiplication or division symbol.	The bar means “given.” Translate the target and condition before choosing a calculation.
Replace the second conditional factor by an overall probability automatically.	Use $P(A) \times P(B\mid A)$ . Only simplify further when the event relationship allows it.
Keep both draw probabilities unchanged without replacement.	Update the remaining target count and total after the first draw.
Complement the condition when using $1$ minus.	$P(A^c\mid B) = 1 – P(A\mid B)$ keeps $B$ fixed. $P(A\mid B^c)$ is a different question.
Say that a $0$ conditioning probability gives conditional probability $0$ .	Division by $0$ is undefined. The basic ratio requires a positive given-event probability.
Accept a conditional probability above $1$ .	Check the data and denominator: a shared group cannot be larger than the given group.
Omit the given group from the interpretation.	State “Among Chess members, $40\%$ belong to Music,” rather than calling $40\%$ the overall Music probability.

Fast reasonableness checks: a conditional probability must lie from $0$ to $1$ . The joint numerator probability cannot exceed the given event probability. If a conditional ratio exceeds $1$ , inspect the event definitions, numerator and denominator.

10 · Try before you reveal

Practice with hints and solutions

Underline the condition first. Write the eligible group or its total before doing any arithmetic.

1. Even given $\text{at least }6$

$1$ ticket is drawn uniformly from numbers $1\text{-}8$ . Write the eligible outcomes and calculate $P(\text{even}\mid \text{at least 6})$ .

Hint for question 1

The condition keeps only $6$ , $7$ and $8$ . Which of those are even?

Solution for question 1

The eligible outcomes are $\{6, 7, 8\}$ ; the target outcomes within that group are $\{6, 8\}$ . The conditional probability is $\frac{2}{3} \approx 0.6667$ .

2. Reverse the ticket question

For the same draw, calculate $P(\text{at least 6}\mid \text{even})$ . Explain why its denominator differs from question 1.

Hint for question 2

The event after the bar is now even. List all $4\text{ even ticket numbers}$ .

Solution for question 2

The given group is $\{2, 4, 6, 8\}$ . The numbers at least $6$ within it are $\{6, 8\}$ , so the probability is $\frac{2}{4} = 0.50$ . The shared outcomes are unchanged, but the condition changes the eligible group from $3$ outcomes $\text{to }4$ .

3. $\text{Greater than }4$ given even

A fair die is $\text{rolled }1\text{ time}$ . Given that its result is even, what is the probability that it is greater than $4$ ?

Hint for question 3

Restrict to $\{2, 4, 6\}$ . “Greater than $4$ ” does not include $4$ .

Solution for question 3

Only $6$ meets the target among the $3\text{ given outcomes}$ . $P(\text{greater than 4}\mid \text{even}) = \frac{1}{3}$ . The whole-space joint probability would instead be $\frac{1}{6}$ .

4. Not Chess given Music

Use the $60$ -student club table. $1$ listed student is selected uniformly. Find $P(\text{not Chess}\mid \text{Music})$ .

Hint for question 4

The given group is the Music Yes column, with $21$ students.

Solution for question 4

$9\text{ of the}$ $21$ Music members are not in Chess. The probability is $\frac{9}{21} = \frac{3}{7} \approx 0.4286$ . This also equals $1 – P(\text{Chess}\mid \text{Music}) = 1 – \frac{4}{7}$ .

5. Use the conditional formula

A model gives $P(A\cap B) = 0.18$ and $P(B) = 0.45$ . Find $P(A\mid B)$ .

Hint for question 5

Divide the joint probability by the probability of the event after the bar.

Solution for question 5

$P(A\mid B) = \frac{0.18}{0.45} = 0.40$ . The condition has positive probability, and the joint probability is no greater than $0.45$ .

6. From conditional to joint

A fictional school model gives $P(\text{Late}) = 0.20$ and $P(\text{Forget notebook}\mid \text{Late}) = 0.35$ . Find the probability of being late and forgetting a notebook.

Hint for question 6

Use the general multiplication rule with Late as the first event.

Solution for question 6

$P(\text{Late}\cap \text{Forget notebook}) = 0.20 \times 0.35 = 0.07 = 7\%$ . The second factor is conditional on Late; the answer is the chance that both events occur.

7. $2\text{ Gold tokens}$ without replacement

Use the bag with $3$ Green and $2$ Gold tokens. Given that the first token is Gold, find the probability that the second is Gold. Then find the probability that both are Gold.

Hint for question 7

After a first Gold token, $1$ Gold token remains among $4$ tokens.

Solution for question 7

$P(\text{second Gold}\mid \text{first Gold}) = \frac{1}{4}$ . $P(\text{both Gold}) = (\frac{2}{5}) \times (\frac{1}{4}) = \frac{1}{10} = 0.10$ . The conditional probability and the joint probability answer different questions.

8. Green then Gold

$2$ tokens are drawn without replacement from the same bag. Find the probability that the first is Green and the second is Gold.

Hint for question 8

After a first Green, both Gold tokens are still in the $4\text{-token}$ bag.

Solution for question 8

$P(\text{first Green}) = \frac{3}{5}$ , and $P(\text{second Gold}\mid \text{first Green}) = \frac{2}{4}$ . The joint probability is $(\frac{3}{5}) \times (\frac{2}{4}) = 0.30$ .

9. Complement the target, not the condition

Let $C$ mean “completes the task” and $O$ mean “uses the online version.” Given $P(C\mid O) = 0.72$ , find $P(C^c\mid O)$ . Does the same information determine $P(C\mid O^c)$ ?

Hint for question 9

Keep $O$ fixed for the complement calculation. Not $O$ is a different given group.

Solution for question 9

$P(C^c\mid O) = 1 – 0.72 = 0.28$ . The information does not determine $P(C\mid O^c)$ , because that asks about users of a different version.

10. Check whether the formula can be used

Consider two separate cases: (i) $P(B) = 0$ and $P(A\cap B) = 0$ ; (ii) a proposed model reports $P(B) = 0.10$ and $P(A\cap B) = 0.20$ . Can either case give a valid value of $P(A\mid B)$ using the ratio formula?

Hint for question 10

Check for division by $0$ , then check whether the joint group can be larger than the given group.

Solution for question 10

(i) Undefined: the ratio is $\frac{0}{0}$ , not $0$ . The condition $P(B) > 0$ fails. (ii) Invalid proposed model: the joint probability cannot exceed $P(B)$ . The ratio $\frac{0.20}{0.10} = 2$ flags incompatible inputs; it is not a valid conditional probability.

11 · Keep the essentials

Quick revision

Read the bar

P(A\mid B)

means

A

given

B

. The right side names the condition.

Identify the eligible groupRestrict to

B

before counting target outcomes.

Keep the shared numeratorUse

A \cap B

: outcomes meeting both conditions.

Choose the denominatorFor counts, use the given group’s total; for probabilities, use

P(B)

Check $P(B) > 0$ The basic conditional ratio is undefined when its denominator is

0

Reverse with care

P(A\mid B)

and

P(B\mid A)

generally use different denominators.

Multiply with a condition

P(A\cap B) = P(A) \times P(B\mid A)

Update without replacementAccount for the first draw when calculating the second-draw chance.

Questions students often ask

Does “given” always mean an earlier event in time?

No. A condition is information used to identify the eligible group. Club membership can be a condition even though the two memberships are recorded at the same time. In sequential draws, an earlier result can also supply the condition.

Does conditioning always change the numerical probability?

No. Conditioning restricts the eligible group to the given event; its target share may equal the overall target share. Use the calculation rather than assuming the value must increase or decrease.

Must I assume independence to use the general multiplication rule?

No. The general rule already uses the appropriate conditional probability. Independence is needed to replace that conditional probability with an unconditional one; that relationship is studied in Topic 2.7.

Can

P(A\mid B)

be larger than

P(A\cap B)

Yes. For $P(B) > 0$ , dividing the joint probability by $P(B)$ can increase its value because $P(B)$ is at most $1$ . In the club table, $P(\text{Music}\mid \text{Chess}) = 0.40$ while $P(\text{Music}\cap \text{Chess}) = 0.20$ .

12 · Show that you understand

Final understanding check

A new original teaching example lists $100$ students and their Art and Science club memberships. $1$ listed student is selected uniformly at random. $A$ means “Art member,” and $S$ means “Science member.”

New original $100$ -student club example. These counts are separate from the earlier Chess/Music table.
Art member?	Science Yes	Science No	Row total
Yes	$18$	$22$	$40$
No	$12$	$48$	$60$
Total	$30$	$70$	$100$

Find $P(A\cap S)$ .
Find $P(A\mid S)$ and interpret it in words.
Find $P(S\mid A)$ . Explain why it differs from $P(A\mid S)$ .
Find $P(A^c\mid S)$ using the same given group.
Find $P(A\mid S^c)$ . Explain why this is not $1 – P(A\mid S)$ .
Use $P(S) \times P(A\mid S)$ to recover the joint probability.
In a separate draw from the $3\text{-Green},2\text{-Gold}$ bag without replacement, calculate $P(\text{first Gold and second Green})$ .

Open the complete final-check solution

$P(A\cap S) = \frac{18}{100} = 0.18$ .
$P(A\mid S) = \frac{18}{30} = 0.60$ . Among Science members in this group, $60\%$ also belong to Art.
$P(S\mid A) = \frac{18}{40} = 0.45$ . It uses the $40$ Art members as its given group, rather than the $30$ Science members.
$P(A^c\mid S) = \frac{12}{30} = 0.40$ , also $1 – 0.60$ .
$P(A\mid S^c) = \frac{22}{70} = \frac{11}{35} \approx 0.3143$ . The given group is the $70$ non-Science members, so it is a different condition.
$P(S) \times P(A\mid S) = (\frac{30}{100}) \times (\frac{18}{30}) = 0.18$ .
First Gold has probability $\frac{2}{5}$ . After a first Gold, $3$ Green tokens remain among $4$ tokens. The joint probability is $(\frac{2}{5}) \times (\frac{3}{4}) = 0.30$ .

Ready to move on? You should be able to name the given group, choose its denominator, reverse a condition carefully, and use a conditional probability in a product. If you used $100$ for every table question, review the given-group totals before continuing.

13 · Keep building

Continue learning

Up next · Topic 2.7Topic 2.7: Independent Events and Unions of Events →

Check when knowing one event does not change another event’s probability. Then calculate probabilities for “ $A$ or $B$ ,” including their shared outcomes.

← Topic 2.5: Mutually Exclusive Events

Review conditional table probabilities · Review the multiplication rule · Back to the lesson overview

0000003998 : Abdad Alam Shamim Alam

This is the bestest institution I have ever come to and I love it very much.

Aliki S

Ali

Sly is him

Jitendra Kumar Kumawat

Raahil Hasan

Tina Mudarres

The classes are amazing and my child learnt so much

alisha gadoya

I have experienced a lot of good things, it has taught me so many things and from B/Cs I have gone to an A! This is wonderful and Mavish’s class is awesome.

smasher 123

It’s sensational my kid went. First he only would get C now it’s all A’s really good and recommend

Mosa Al- Samaraie

AMAZING PRICES GREAT TEACHERS STRAIGHT FORWARD LEARNING STEADY PACE IN TUTORING

Cael Dagnelie

I had a great experience learning math with Ms. Mavish. She explains complex topics in a very clear and simple way, which made it easier for me to understand and enjoy the subject. Her patience and dedication really stood out, and she always made sure that everyone in the class was keeping up. I especially appreciated how approachable she was — I never felt afraid to ask questions, and she was always willing to help. Thanks to her teaching, my confidence in math has grown a lot. I’m really thankful for the effort she puts into every lesson!