4.5 Conditional Probability and Expected Value
Key Takeaways
- Conditional probability P(A|B) is the probability of A given that B has already occurred: P(A and B) / P(B) when P(B) > 0.
- Events are independent if P(A|B) = P(A); otherwise the first outcome changes the probability of the second.
- Without replacement, update both numerator and denominator after each draw — successive events are dependent.
- Expected value is the weighted average of outcomes: sum of (value × probability) for all possibilities.
- Two-way tables organize joint and marginal counts and are the fastest tool for many Praxis conditional probability items.
Why This Section Matters
Probability items on Praxis Mathematics (5165) go beyond coin flips. Expect two-way tables, conditional probability, without-replacement draws, and expected value in context (games, fundraisers, diagnostic tests). You must read whether the problem says "given that" or "and" — that one phrase changes the denominator.
Basic Probability Rules
For equally likely outcomes in a finite sample space:
P(event) = favorable outcomes / total outcomes
For compound events:
- P(A and B) — both happen (multiply when independent; adjust when dependent).
- P(A or B) — at least one happens: P(A) + P(B) − P(A and B).
Always check whether the problem states with replacement (independent trials) or without replacement (dependent trials).
Conditional Probability
P(A | B) means the probability of A given that B occurred.
P(A | B) = P(A and B) / P(B), provided P(B) > 0.
The denominator is not always the full sample space — it is the conditioning event B.
Worked Example: Two-Way Table
In a class: 12 students play a sport, 8 do not. Among sport players, 9 passed a quiz; among non-players, 4 passed.
| Passed | Did not pass | Total | |
|---|---|---|---|
| Plays sport | 9 | 3 | 12 |
| Does not play | 4 | 4 | 8 |
| Total | 13 | 7 | 20 |
If a passing student is chosen at random, what is the probability the student plays a sport?
This is P(sport | passed) = 9 / 13, not 9/20. The condition "chosen from those who passed" restricts the denominator to 13 passing students.
By contrast, P(passed | sport) = 9/12 = 3/4 — a different question with a different denominator.
Independent vs Dependent Events
Events A and B are independent if knowing B does not change the probability of A:
P(A | B) = P(A).
Drawing with replacement often produces independence. Drawing without replacement changes counts for the second draw.
Worked Example: Without Replacement
A box has 5 red and 3 blue balls. Two balls are drawn without replacement. Find P(both red).
First draw red: 5/8.
Second draw red given first was red: 4/7 (one fewer red, one fewer total).
P(both red) = (5/8)(4/7) = 20/56 = 5/14.
Common error: using (5/8)(5/8) as if the draws were independent — that treats the situation with replacement.
Worked Example: Exactly k Successes in n Trials
A fair coin is tossed 3 times. Find P(exactly 2 heads).
List favorable sequences: HHT, HTH, THH — 3 outcomes.
Total outcomes = 2³ = 8 equally likely sequences.
P(exactly 2 heads) = 3/8.
For small n, listing remains efficient on the exam; recognize the binomial pattern when appropriate.
Expected Value
The expected value E(X) of a random variable is the long-run average if the chance process is repeated many times:
E(X) = Σ [outcome × P(outcome)]
Worked Example: Game Payout
A game pays $10 with probability 0.2 and $2 with probability 0.8.
E(X) = 10(0.2) + 2(0.8) = 2 + 1.6 = $3.60.
A single play may yield $10 or $2, but $3.60 is the fair long-run average payout per game. Fundraiser items may ask whether a price of $5 per play favors the school — compare $5 to the expected payout.
Organizing Information
| Tool | Use when |
|---|---|
| Two-way table | Categories cross-classified (sport vs pass/fail) |
| Tree diagram | Sequential events, especially without replacement |
| Venn diagram | "And / or / not" relationships among events |
Praxis Traps
- Wrong denominator on conditional problems — read "given that" carefully.
- Treating without replacement as independent — update counts after each draw.
- Expected value as a guaranteed outcome — it is an average, not a certain payment.
- Adding probabilities that are not mutually exclusive without subtracting overlap.
Teaching Connection
When students confuse P(A|B) with P(B|A), have them write both denominators in words: "out of everyone who passed" versus "out of everyone who plays a sport." That language fix resolves many errors that look like arithmetic mistakes on Praxis task-of-teaching items.
Complement and "At Least One" Problems
The complement rule states P(not A) = 1 − P(A). For "at least one success in several trials," it is often faster to compute 1 − P(none).
Worked Example: At Least One Head
A fair coin is tossed 3 times. Find P(at least one head).
P(no heads) = P(all tails) = (1/2)³ = 1/8.
P(at least one head) = 1 − 1/8 = 7/8.
Listing {H**, *H, ***H} confirms seven favorable outcomes out of eight.
Fair Games and Expected Value Decisions
A carnival game is fair when the price to play equals the expected payout. If expected payout is $3.60 but the game costs $5, players lose money on average — the operator profits. Praxis may ask whether a fundraiser is favorable to the school or to the player; compare cost to E(X) directly.
| Outcome | Probability | Contribution to E(X) |
|---|---|---|
| $10 prize | 0.2 | $2.00 |
| $2 prize | 0.8 | $1.60 |
| Expected payout | — | $3.60 |
In a class, 12 students play a sport and 8 do not. Of sport players, 9 passed a quiz; of non-players, 4 passed. If one student who passed is selected at random, what is the probability the student plays a sport?
A box contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. What is the probability both are red?
A game pays $10 with probability 0.2 and $2 with probability 0.8. What is the expected payout?