الإحصاء (بالإنجليزية) الصف الثالث الثانوي، الأسبوع 13: Using the formula both ways: finding P(A ∩ B) = P(A | B) × P(B) and reading two-way tables

Conditional probability

الأسبوع 13 · الاحتمال الشرطي: فضاء العينة المصغّر و⁦P(A | B) = P(A ∩ B) ÷ P(B)⁩ وقاعدة الضرب والجداول التوافقية والشجرة وشكل فن · 3 دروس

Using the formula both ways: finding P(A ∩ B) = P(A | B) × P(B) and reading two-way tables

Statistics Sec 3 Term 1, Unit 3, Lesson 2

GoalUse the formula the other way, P(A ∩ B) = P(A | B) × P(B), and read conditional probabilities from a two-way table.

New words

multiplication rule · P(A ∩ B) = P(A | B) × P(B) = P(B | A) × P(A)two-way table · a table that sorts the same people or objects by two featuresrow and column totals · the totals at the edges, which give the probabilities of single events

Explanation

Rearranging P(A | B) = P(A ∩ B) ÷ P(B) gives the multiplication rule P(A ∩ B) = P(A | B) × P(B). In words: for both events to happen, B must happen first, and then A must happen in the world where B has happened. By symmetry P(A ∩ B) = P(B | A) × P(A) as well, and the two products must be equal. A two-way table is the most direct place to use these ideas. Here 200 pupils are sorted by grade and by whether they bring lunch. P(brings lunch) = 120/200 = 0.6. But P(brings lunch | Grade 12) = 45/65 = 9/13, about 0.69: given Grade 12, the reduced sample space is the Grade 12 column, 65 pupils. And P(Grade 12 | brings lunch) = 45/120 = 3/8 = 0.375, because now the reduced sample space is the first row. The same cell, 45, is divided by two different totals, and that is why the two conditional probabilities differ.

	Grade 10	Grade 11	Grade 12	Total
Bring lunch	40	35	45	120
Buy lunch	30	30	20	80
Total	70	65	65	200

Worked examples

From the table: P(buys lunch | Grade 11) = 30/65 = 6/13, using the Grade 11 column. P(Grade 11 | buys lunch) = 30/80 = 3/8, using the “buys lunch” row. Both use the same cell, 30, and different totals.

The multiplication rule. On a school trip day the probability that the morning is cloudy is 0.5, and the probability of rain given a cloudy morning is 0.3. Then P(cloudy and rain) = 0.5 × 0.3 = 0.15. The probability of rain is at least 0.15, because the cloudy-and-rain outcomes are part of all the rainy ones.

Filling in. P(A) = 0.6, P(B | A) = 0.5 and P(B) = 0.45. Then P(A ∩ B) = 0.5 × 0.6 = 0.3; P(A | B) = 0.3 ÷ 0.45 = 2/3; P(A ∪ B) = 0.6 + 0.45 − 0.3 = 0.75; and P(B′ | A) = 1 − 0.5 = 0.5, because given A the outcomes B and B′ share the whole.

Class activity

Ask 20 people two yes/no questions, for example “Do you wear a watch?” and “Are you left-handed?”, and make a 2 × 2 table with totals. Compute P(A | B) and P(B | A) from it and show that they use the same cell with different totals.

Practice

P(A) = 0.5 and P(B | A) = 0.6. Find P(A ∩ B).

In the sports-club table, find P(swimming | 16 or over).

	Under 16	16 or over	Total
Swimming	30	25	55
Running	20	45	65
Total	50	70	120

In the same table, find P(under 16 | running).

	Under 16	16 or over	Total
Swimming	30	25	55
Running	20	45	65
Total	50	70	120

P(A ∩ B) = 0.24, P(A) = 0.4 and P(B) = 0.5. Find P(B | A) and P(A | B).

P(A | B) = 0.35. Find P(A′ | B) and explain.