Conditional probability

About this lesson

Integrated Math II › Applications of probability

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Conditional probability

Warm-up

Three quick questions from earlier lessons.

Discover

Of 100 students, 40 take French; 10 of those also take Spanish. Among the French students, what is the probability of taking Spanish?

  1. 1040\tfrac{10}{40}
  2. 10100\tfrac{10}{100}

Key idea

Conditioning shrinks the sample space.

P(A∣B)=P(A and B)P(B)P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}: the probability of AA among the outcomes in BB. In general P(A∣B)≠P(B∣A)P(A \mid B) \ne P(B \mid A).

Worked example

P(rain) = 0.3, P(late) = 0.2, P(rain and late) = 0.12.

P(late∣rain)=0.120.3=0.4P(\text{late} \mid \text{rain}) = \tfrac{0.12}{0.3} = 0.4
Divide by P(rain).
P(rain∣late)=0.120.2=0.6P(\text{rain} \mid \text{late}) = \tfrac{0.12}{0.2} = 0.6
Divide by P(late).
Two different numbers
Different conditions.

Your turn, guided

Your turn: P(A and B) = 0.1, P(B) = 0.4. Find P(A | B).

0.250.25
0.1 ÷ 0.4.

Practice ★

  1. P(A) = 0.60, P(B) = 0.22 and P(A and B) = 0.5. Find P(A | B).
  2. P(A) = 0.55, P(B) = 0.33 and P(A and B) = 0.13. Find P(A | B).
  1. A screening test was given to 316 people: 47 had the disease and tested positive, 8 had the disease and tested negative, 34 did not have it and tested positive, 227 did not have it and tested negative. Find P(disease | positive).
  2. A screening test was given to 256 people: 33 had the disease and tested positive, 7 had the disease and tested negative, 21 did not have it and tested positive, 195 did not have it and tested negative. Find P(disease | positive).

Watch out!

Practice ★★

  1. A test detects a disease in 87% of people who have it. Which statement says the same thing?
    1. P(positive | disease) = 0.87
    2. P(disease | positive) = 0.87
    3. P(disease and positive) = 0.87

Exit ticket

Three questions to check you have got it.

  1. A screening test was given to 316 people: 47 had the disease and tested positive, 8 had the disease and tested negative, 34 did not have it and tested positive, 227 did not have it and tested negative. Find P(disease | positive).
  2. A screening test was given to 256 people: 33 had the disease and tested positive, 7 had the disease and tested negative, 21 did not have it and tested positive, 195 did not have it and tested negative. Find P(disease | positive).
  3. A screening test was given to 239 people: 57 had the disease and tested positive, 3 had the disease and tested negative, 29 did not have it and tested positive, 150 did not have it and tested negative. Find P(disease | positive).

Summary card

Conditional probability

P(A∣B)=P(A and B)/P(B)P(A \mid B) = P(A \text{ and } B) / P(B).

Watch out: P(A∣B)P(A \mid B) and P(B∣A)P(B \mid A) are different.

End of the lesson

Next lesson: Independence and the multiplication rule Back to the chapter

Answers: Conditional probability

A

  1. 522\frac{5}{22}
  2. 1333\frac{13}{33}

B

  1. 4781\frac{47}{81}
  2. 1118\frac{11}{18}

C

  1. P(positive | disease) = 0.87

D

  1. 4781\frac{47}{81}
  2. 1118\frac{11}{18}
  3. 5786\frac{57}{86}

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