Week 13 · Conditional probability: the reduced sample space, P(A | B) = P(A ∩ B) ÷ P(B), the multiplication rule, two-way tables, tree and Venn diagrams · 3 lessons
3
Two draws without replacement on a tree diagram, a Venn diagram of two groups, and why P(A | B) is not P(B | A)
Statistics Sec 3 Term 1, Unit 3, Lesson 2
GoalUse a tree diagram for two draws without replacement, read conditional probabilities from a Venn diagram, and explain why P(A | B) differs from P(B | A).
New words
without replacement · the first item drawn is not put backtree diagram · branches carry probabilities; multiply along a pathP(A | B) ≠ P(B | A) · reversing the condition changes the question
Explanation
A bag holds 4 green and 3 yellow counters. Two are drawn one after the other without putting the first back. On the tree (G green, Y yellow) the first branches carry 4/7 and 3/7. After a green counter is taken, 6 remain, 3 green and 3 yellow, so the second branches are 3/6 and 3/6; these are conditional probabilities, P(second | first). The probability of a whole path is the product along it, which is the multiplication rule: P(G then G) = 4/7 × 3/6 = 2/7. The same tree answers conditional questions: P(first G | second Y) = P(G then Y) ÷ P(second Y) = (2/7) ÷ (2/7 + 1/7) = 2/3. A Venn diagram or a table of counts does the same job. P(A | B) is the overlap divided by the whole of B, and P(B | A) is the same overlap divided by the whole of A. Since the two sets usually have different sizes, the two answers usually differ.
Worked examples
The four paths of the tree: GG = 4/7 × 3/6 = 2/7; GY = 4/7 × 3/6 = 2/7; YG = 3/7 × 4/6 = 2/7; YY = 3/7 × 2/6 = 1/7. They add up to 1. One counter of each colour: GY and YG, so 2/7 + 2/7 = 4/7. And P(second Y | first G) = 3/6 = 1/2 is read straight from the branch.
A Venn diagram. Of 60 pupils, 36 borrow library books (A), 24 use the computer room (B) and 12 do both. Then P(A) = 0.6, P(B) = 0.4 and P(A ∩ B) = 0.2. P(A | B) = 0.2 ÷ 0.4 = 0.5, but P(B | A) = 0.2 ÷ 0.6 = 1/3. The four regions are 0.4, 0.2, 0.2 and 0.2 (24, 12, 12 and 12 pupils).
When one event lies inside the other. A box has 20 cards: 10 show a rectangle, of which 3 are squares, 6 show a circle and 4 a triangle. Let R = “rectangle” and Q = “square”, so Q ⊂ R. P(Q | R) = (3/20) ÷ (10/20) = 0.3, but P(R | Q) = (3/20) ÷ (3/20) = 1: a square card is certainly a rectangle.
Class activity
Take 4 red pens and 3 blue pens (or any seven objects of two kinds). Draw two without looking and without putting the first back, record the colours, and put them back. Repeat 20 times. What fraction of your draws were both red? Compare it with 4/7 × 3/6 = 2/7 ≈ 0.29.
Practice
1
A bag has 6 white and 3 black counters. Two are drawn without replacement. Find P(both white), P(white then black) and P(one of each colour).
