Conditional Probability

11.6: Conditional Probability. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 6 of 63 parts Practice Flashcards

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Condition on a group

Objective: Compute a probability inside a specified subgroup.

Prerequisite skills: simple probability and two-way tables. P(A|B) means the probability of A among outcomes where B is true. Restrict the sample space to B first. The denominator must be P(B) or the count of B and must be positive. Common error: using the full sample after a condition is imposed. Restrictions and interpretation: A conditional probability changes the reference population; independence must be established, not assumed.

Formulas

$P(A|B)=P(A∩B)/P(B), P(B)>0$

Worked examples

Try itA survey has 288 people. Of 120 who take a bus, 48 also ride a bike. Given that a person takes a bus, what is the probability they also ride a bike?Show answer

The condition narrows the population to the 120 bus riders. Of those, 48 ride bikes. Probability=48/120=2/5=0.4. Therefore the requested answer is 0.4.

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Condition on a group

Objective: Compute a probability inside a specified subgroup.

Formulas

$P(A|B)=P(A∩B)/P(B), P(B)>0$

Worked examples

Try itA survey has 288 people. Of 120 who take a bus, 48 also ride a bike. Given that a person takes a bus, what is the probability they also ride a bike?Show answer

The condition narrows the population to the 120 bus riders. Of those, 48 ride bikes. Probability=48/120=2/5=0.4. Therefore the requested answer is 0.4.

Without-replacement probability

Objective: Update counts after a first selection.

When an item is selected without replacement, both total count and the count of the selected category can change. A conditional second-draw probability is computed in the reduced collection. Common error: keeping the original denominator after removing an item. Restrictions and interpretation: A conditional probability changes the reference population; independence must be established, not assumed.

Formulas

P(second red | first red)=(r−1)/(N−1)

Worked examples

Try itA bag contains 25 red and 27 blue beads. A first bead is red and is not replaced. What is the probability the second bead is red?Show answer

After the known red draw, 24 red beads remain out of 51 total. Thus the probability is 24/51. Therefore the requested answer is 8/17.

Test independence

Objective: Compare conditional and unconditional probabilities.

Events are independent when conditioning on one leaves the probability of the other unchanged. Equivalently, P(A∩B)=P(A)P(B). Independence differs from mutual exclusivity: disjoint positive-probability events are not independent. Common error: equating independence with no overlap. Restrictions and interpretation: A conditional probability changes the reference population; independence must be established, not assumed. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

$independent ⇔ P(A|B)=P(A) when P(B)>0$

Worked examples

Try itAmong 480 people, 240 have A, 192 have B, and 96 have both. Are A and B independent in this distribution?Show answer

P(A)=240/480=1/2. P(A|B)=96/192=1/2. The matching probabilities establish independence; overlap does not prevent it. Therefore the requested answer is Yes, because P(A|B) = P(A).

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