الأسبوع 13 · الاحتمال الشرطي: فضاء العينة المصغّر وP(A | B) = P(A ∩ B) ÷ P(B) وقاعدة الضرب والجداول التوافقية والشجرة وشكل فن · 3 دروس
1
From P(A) to P(A | B): knowing that B happened shrinks the sample space, and the formula P(A | B) = P(A ∩ B) ÷ P(B)
Statistics Sec 3 Term 1, Unit 3, Lesson 2
GoalExplain why knowing that B occurred shrinks the sample space, and compute P(A | B) = P(A ∩ B) ÷ P(B).
New words
conditional probability P(A | B) · the probability of A when we know that B has occurredgiven that · the words that introduce the conditionreduced sample space · the event B itself, once we know it occurred
Explanation
Sometimes we learn something about the result before we look at it. A die is thrown and a friend says the number is even. The chance of “more than 3” was 3/6 = 1/2 before the news. Now only 2, 4 and 6 are possible, and two of them (4 and 6) are more than 3, so the chance becomes 2/3. Knowing that B occurred has reduced the sample space from S to B. In general P(A | B), read “the probability of A given B”, is the share of B that also lies in A: P(A | B) = n(A ∩ B) ÷ n(B) = P(A ∩ B) ÷ P(B), where P(B) > 0. Dividing by P(B) is the step that makes B the new whole, so that P(B | B) = 1. The two forms agree because the common factor 1 ÷ n(S) cancels. Read the wording with care: the event after “the probability of” is the one we want, and the event after “given that” is the condition. The table lists the six outcomes of the die.
1
2
3
4
5
6
even (B)
no
yes
no
yes
no
yes
more than 3 (A)
no
no
no
yes
yes
yes
both (A ∩ B)
no
no
no
yes
no
yes
Worked examples
The die above, with A = “more than 3” and B = “even”. Reduced sample space B = {2, 4, 6}; A ∩ B = {4, 6}. So P(A | B) = 2/3. With the formula: P(A ∩ B) ÷ P(B) = (2/6) ÷ (3/6) = 2/3. The condition raised the chance from 1/2 to 2/3.
Reduced sample space: first die even (sums shown)
1
2
3
4
5
6
2
3
4
5
6
7
8
4
5
6
7
8
9
10
6
7
8
9
10
11
12
Two dice. Find P(the sum is 8 given that the first die is even). The first die is 2, 4 or 6, so the reduced sample space has 3 × 6 = 18 pairs, shown below. The sum 8 appears in 3 of them, (2, 6), (4, 4) and (6, 2). So the probability is 3/18 = 1/6. Without the condition P(sum 8) = 5/36, which is a little smaller.
Three coins. Find P(exactly two heads given that there is at least one head). “At least one head” keeps 7 of the 8 outcomes (all but TTT). Among them HHT, HTH and THH have exactly two heads. So the answer is 3/7. By the formula: (3/8) ÷ (7/8) = 3/7. The unconditional P(exactly two heads) is 3/8, so the condition changed the answer.
Class activity
Write the numbers 1 to 12 on slips. Suppose someone tells you only that the number drawn is prime. List the reduced sample space, find P(odd | prime) and compare it with P(odd) = 1/2. Explain in a sentence why the two answers are different.
Practice
1
A die is thrown. Let A = “5 or 6” and B = “greater than 2”. Find P(A | B).
B = {3, 4, 5, 6} and A ∩ B = {5, 6}, so P(A | B) = 2/4 = 1/2.
2
A spinner has 12 equal sectors numbered 1 to 12. Find P(a multiple of 4 | even).
Even: {2, 4, 6, 8, 10, 12}; multiples of 4 among them: 4, 8, 12. So 3/6 = 1/2 (compare 3/12 = 1/4 without the condition).
3
Two coins are tossed. Find P(two heads | at least one head).
At least one head: HH, HT, TH. Only HH has two heads, so 1/3.
4
P(A ∩ B) = 0.18 and P(B) = 0.45. Find P(A | B).
P(A | B) = 0.18 ÷ 0.45 = 0.4.
5
“Of the pupils who arrived late, what fraction were girls?” Write this as P( | ) and say which event is the condition.
P(girl | late). The condition is “late”: we look only at the late pupils.
