Probability and Conditional Probability Pattern - Conditional Probability

Finding the probability of an event given that another event already happened

Conditional probability is one of the most commonly tested concepts in SAT data analysis. The question tells you a condition has already occurred, and you must find the probability of another event within that restricted group.

The Core Idea

$P(A \mid B) = \dfrac{\text{number in both A and B}}{\text{total in B}}$

The key word is "given." When you see "given that the item is B," the denominator shrinks from the grand total to only the items in category B. Everything outside B is irrelevant.

What This Looks Like — Two-Way Table

A used car dealership has the following inventory of 155 vehicles:

	Black	White	Silver	Other	Total
Sedan	20	15	10	5	50
SUV	25	20	15	10	70
Truck	15	10	5	5	35
Total	60	45	30	20	155

If a vehicle is selected at random from the inventory, what is the probability that the vehicle is an SUV, given that its color is not black?

A) $\dfrac{9}{31}$
B) $\dfrac{9}{19}$
C) $\dfrac{9}{14}$
D) $\dfrac{4}{9}$

Step 1 — Restrict to the "given" group. The condition is "not black," so the new total is all non-black vehicles: $155 - 60 = 95$.
Step 2 — Count favorable within that group. Non-black SUVs: $20 + 15 + 10 = 45$.
Step 3 — Form the conditional fraction: $P(\text{SUV} \mid \text{not black}) = \dfrac{45}{95} = \dfrac{9}{19}$
Answer: B

The Three Classic Wrong Answers in Table Problems

1. Using the grand total as the denominator: $\dfrac{45}{155} = \dfrac{9}{31}$ — this ignores the "given" restriction entirely.

2. Flipping the conditional: $\dfrac{45}{70} = \dfrac{9}{14}$ — this answers "P(not black | SUV)" instead of "P(SUV | not black)." The denominator should be the "given" group, not the "event" group.

3. Restricting too much: $\dfrac{20}{45} = \dfrac{4}{9}$ — this only considers white vehicles instead of all non-black vehicles.

What This Looks Like — Word Problem (No Table)

A factory has two assembly lines, Line A and Line B. Line A operates in $4$ production runs, each producing $10$ premium and $30$ standard models. Line B operates in $7$ runs, each producing $15$ premium and $25$ standard models. If a smartphone is selected at random, what is the probability it came from Line B, given that it is a premium model?

A) $\dfrac{3}{5}$
B) $\dfrac{21}{88}$
C) $\dfrac{15}{440}$
D) $\dfrac{21}{29}$

Step 1 — Compute totals.
Line A premium: $4 \times 10 = 40$
Line B premium: $7 \times 15 = 105$
Total premium (the "given" group): $40 + 105 = 145$

Step 2 — Form the conditional fraction:
$P(\text{Line B} \mid \text{premium}) = \dfrac{105}{145} = \dfrac{21}{29}$
Answer: D

Gotcha: Per-Unit vs. Total

A sneaky trap in word problems with "runs" or "shipments" is using the per-unit numbers instead of the totals. In the factory problem:

Using per-run values gives $\dfrac{15}{10 + 15} = \dfrac{15}{25} = \dfrac{3}{5}$ — wrong, because Line A has 4 runs while Line B has 7. The number of runs changes the weighting.

Always multiply the per-unit count by the number of units before forming your fraction.

How to Identify "Given" in a Question

The SAT phrases the condition in several ways:

• "given that the item is X"
• "among those who are X"
• "of the items that are X"
• "if it is known that the item is X"

All of these mean the same thing: restrict your denominator to category X.

What to Do on Test Day

• Find the "given" group first — this becomes your denominator. Cross out everything else mentally.
• Then count favorable outcomes within that group — this is your numerator.
• Never use the grand total as the denominator in a conditional probability question. The whole point is that the sample space is restricted.
• Watch for the "flip": P(A|B) $\neq$ P(B|A). Make sure your denominator matches the "given" condition, not the event you're looking for.
• Key formula: $P(A \mid B) = \dfrac{\text{count in both A and B}}{\text{total count in B}}$
• In word problems with rates/runs/shipments, always multiply to get totals before dividing.