Tree diagrams and reverse conditioning

About this lesson

Statistics (AP-aligned) › Probability

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Tree diagrams and reverse conditioning

Warm-up

Three quick questions from earlier lessons.

Discover

1% of people have a condition. A test is positive for 95% of them, and for 5% of healthy people. Of 10,000 people, about how many test positive in total?

  1. About 590
  2. About 5,000

Key idea

A tree organizes conditional probabilities.

First branches: P(A)P(A) and P(not A)P(\text{not } A); second branches: conditional probabilities. Multiply along a path; add the paths that end in BB to get P(B)P(B). Then P(A∣B)=P(A and B)P(B)P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)} (reverse conditioning, Bayes' rule).

Worked example

With the numbers above: P(condition | positive).

0.01×0.95=0.00950.01 \times 0.95 = 0.0095
Path: condition, positive.
P(+)=0.0095+0.99×0.05=0.059P(+) = 0.0095 + 0.99 \times 0.05 = 0.059
Both paths to positive.
P(condition∣+)=0.00950.059≈0.16P(\text{condition} \mid +) = \tfrac{0.0095}{0.059} \approx 0.16
Divide.

Your turn, guided

Your turn: P(A) = 0.2, P(B | A) = 0.5, P(B | not A) = 0.25. Find P(B).

0.30.3
0.2 × 0.5 + 0.8 × 0.25.

Practice ★

  1. 12% of a population has a condition. A test is positive for 86% of people with the condition and for 11% of people without it. Find P(positive).
  2. 14% of a population has a condition. A test is positive for 94% of people with the condition and for 8% of people without it. Find P(positive).
  1. 22% of a population has a condition. A test is positive for 81% of people with the condition and for 12% of people without it. Find P(condition | positive).

Watch out!

Practice ★★

  1. A disease affects 1 in 1,000 people. A test is 93% accurate for both sick and healthy people. Your test is positive. What is most likely?
    1. You are still more likely healthy: false positives from the many healthy people outnumber the true positives
    2. You are about 93% likely to be sick
    3. You are certainly sick
  1. Boss problem. 5% of people have a disease. A test is positive for 90% of people with the disease and for 7% of people without it.

Exit ticket

Three questions to check you have got it.

  1. 12% of a population has a condition. A test is positive for 86% of people with the condition and for 11% of people without it. Find P(positive).
  2. 14% of a population has a condition. A test is positive for 94% of people with the condition and for 8% of people without it. Find P(positive).
  3. 28% of a population has a condition. A test is positive for 95% of people with the condition and for 14% of people without it. Find P(positive).

Summary card

Trees

  • Multiply along paths; add paths to the same end.
  • P(A∣B)=P(A and B)/P(B)P(A \mid B) = P(A \text{ and } B)/P(B).

Watch out: with rare events, most positives can be false.

End of the lesson

Next lesson: Long-run frequency and simulation Back to the chapter

Answers: Tree diagrams and reverse conditioning

A

  1. 15\frac{1}{5}
  2. 5012500\frac{501}{2500}

B

  1. 99151\frac{99}{151}

C

  1. You are still more likely healthy: false positives from the many healthy people outnumber the true positives

D

  1. 90223\frac{90}{223}

E

  1. 15\frac{1}{5}
  2. 5012500\frac{501}{2500}
  3. 9172500\frac{917}{2500}

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