Conditional probability from trees
Reading conditional prob: P(B|A) is the probability on second-stage branch GIVEN first stage was A.
Worked example
Factory: Machine A makes 60% of items, 2% defective.
Machine B makes 40%, 3% defective.
Find P(defective|Machine A).
Solution
- Tree: First stage: A (0.6) or B (0.4)
- From A: defective (0.02) or good (0.98)
- P(defective|A)=0.02 (second-level branch)
Final answer
0.02. Branch probability in conditional state.
IB-style question — reverse conditional [6 marks]
A shop is supplied bread by two bakeries. Bakery X supplies 70% of the loaves and Bakery Y supplies the other 30%. 4% of Bakery X's loaves are stale, while 10% of Bakery Y's loaves are stale. A loaf is chosen at random.
(a) Find the probability that the loaf is stale.
(b) Given that the chosen loaf is stale, find the probability that it came from Bakery Y.
Step by step
- (a) A loaf is stale along two tree paths: from X, or from Y. Find each path then add.
- Substitute the branch probabilities.
- (b) Use the conditional rule. The intersection 'Y and stale' is the Bakery-Y path from part (a).
- Evaluate.
Final answer
(a) P(stale) = 0.058. (b) P(Y | stale) = 0.030/0.058 = 0.517 (3 s.f.).
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"Given that" the result: work backwards
Tap to the last step: one path on top, all the paths that give the result underneath.
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Which spinner?
In a game, Mia rolls a fair die. If it shows 6 she spins spinner X, which wins with probability 1/2. Otherwise she spins spinner Y, which wins with probability 1/5.
(a) Find the probability that Mia wins.
(b) Given that Mia wins, find the probability that she rolled a 6.
Solution
- (a) Add the two winning paths.
- (b) Top: the winning path through 6. Bottom: all winning paths (part a).
Final answer
(a) 1/4 = 0.25. (b) 1/3 ≈ 0.333.
"Given that they match"?
Bottom = every matching path.
Example: 7 white and 3 black socks, two taken. P(both black | same colour) = (3/10 × 2/9) ÷ (7/10 × 6/9 + 3/10 × 2/9) = 1/8.
Which is the top?
The path that has BOTH things: the result you were given and the one asked about. It is always one of the bottom paths.
Exam tip: The bottom of the fraction is usually your answer to the previous part: write "÷ (a)" and use the unrounded value.
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Conditional probability from a table or a Venn
"Given" shrinks the total: "Given that the person is under 18" means: look only at the under-18 row. Its total is the bottom of the fraction.
| Android | iPhone | Total | |
|---|---|---|---|
| Under 18 | 34 | 26 | 60 |
| 18 or over | 61 | 79 | 140 |
| Total | 95 | 105 | 200 |
Restrict to the row
A phone shop surveyed 200 customers (table above). A customer is chosen at random. Given that the customer is under 18, find the probability that they use an iPhone.
Solution
- Only the under-18 row counts: 60 customers, 26 with an iPhone.
Final answer
26/60 ≈ 0.433.
An unknown from a given conditional
In a Venn diagram, 8 people are in M only and x people are in both D and M. Given that a person is in M, the probability that they are also in D is 0.6. Find x.
Solution
- Given M: the bottom is everyone in M, which is x + 8. The top is the overlap, x.
- Multiply out and solve.
Final answer
x = 12.
Exam tip: Name the "given" group before you divide: "restricted to the 60 under 18" shows the examiner the right denominator.
Test results: false positives and expected numbers
| Result | Meaning | Path on the tree |
|---|---|---|
| False positive | Positive test, but no condition | no condition → positive |
| False negative | Negative test, but has the condition | condition → negative |
| Incorrect result | Either of the two above | add both paths |
The false positives come from the large "no condition" branch: watch how big that path is.
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A screening test
3% of a population have a condition. A test is positive for 95% of people with the condition and for 4% of people without it.
(a) Find the probability that a person tests positive.
(b) Given that a person tests positive, find the probability that they have the condition.
(c) Find the expected number of incorrect results in 2000 tests.
Solution
- (a) Add the two positive paths.
- (b) Top: condition and positive. Bottom: all positives.
- (c) Incorrect = false negative + false positive.
- Expected number = number of tests × probability.
Final answer
(a) 0.0673. (b) ≈ 0.423. (c) about 80.6 incorrect results.
Why is (b) so low?: Most people do not have the condition, so the 4% false positives from the large healthy group outnumber the true positives.
Exam tip: An expected number does not have to be a whole number: give it to 3 significant figures unless told otherwise.