الأسبوع 14 · الأحداث المستقلة وغير المستقلة: اختبار P(A ∩ B) = P(A) × P(B) والسحب مع الإرجاع وبدونه والاستقلال مقابل التنافي · 3 دروس
3
Dependent events and drawing without replacement, independent against mutually exclusive, and the complements of independent events
Statistics Sec 3 Term 1, Unit 3, Lesson 3
GoalHandle dependent events and draws without replacement, tell independent from mutually exclusive, and see that complements of independent events are independent.
New words
dependent draws · draws without replacement: the first draw changes the bagexclusive and independent · two different ideas: A ∩ B = ∅ against P(A ∩ B) = P(A)P(B)complements · if A and B are independent, so are A′ and B′
Explanation
For dependent events the multiplication rule still holds, but with a conditional probability: P(A ∩ B) = P(A) × P(B | A). Drawing without replacement is the standard case. From a bag of 5 red and 2 blue counters, P(two red) = 5/7 × 4/6 = 10/21, because after one red counter is taken only 4 red remain among 6; with replacement it was 25/49. Do not confuse independent with mutually exclusive. If A and B are exclusive and both have positive probability, then P(A ∩ B) = 0 while P(A)P(B) > 0, so they are dependent: knowing that A happened tells you that B did not. A coin gives a clean example, since H and T are exclusive and P(H ∩ T) = 0 ≠ 1/4. Overlapping events can be independent, like “even” and “a multiple of 3” on a die. Finally, if A and B are independent then so are A′ and B′: P(A′ ∩ B′) = 1 − P(A ∪ B) = 1 − P(A) − P(B) + P(A)P(B) = (1 − P(A))(1 − P(B)).
Worked examples
The tree for 5 red and 2 blue without replacement: RR = 5/7 × 4/6 = 10/21; RB = 5/7 × 2/6 = 5/21; BR = 2/7 × 5/6 = 5/21; BB = 2/7 × 1/6 = 1/21. They add up to 1. One of each colour: 10/21 ≈ 0.48, compared with 20/49 ≈ 0.41 when the counters are put back.
Events
P(A ∩ B)
P(A) × P(B)
Exclusive?
Independent?
coin: H and T
0
1/4
yes
no
die: even and odd
0
1/4
yes
no
die: even and a multiple of 3
1/6
1/6
no
yes
die: even and at most 3
1/6
1/4
no
no
Independent against exclusive. The table sorts four pairs. Only the pair “even and a multiple of 3” is independent, although the events overlap; the exclusive pairs are dependent, and the last pair is neither.
Complements. Let A and B be independent with P(A) = 0.3 and P(B) = 0.4. Then P(A ∩ B) = 0.12, P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58, and P(A′ ∩ B′) = 1 − 0.58 = 0.42 = 0.7 × 0.6 = P(A′) × P(B′). Also P(A′ ∩ B) = 0.4 − 0.12 = 0.28 = 0.7 × 0.4.
Class activity
Put 5 red and 2 blue objects in a bag. Draw two without replacement, 20 times, and count the draws with two red; compare with 10/21 ≈ 0.48. Then repeat with replacement and compare with 25/49 ≈ 0.51. Which of the two counts is closer to its prediction? Is that always so?
Practice
1
A bag holds 5 green and 4 yellow counters. Two are drawn without replacement. Find P(two green), P(two yellow) and P(one of each colour).
GG = 5/9 × 4/8 = 5/18; YY = 4/9 × 3/8 = 1/6; one of each = 1 − 5/18 − 3/18 = 10/18 = 5/9.
2
A and B are exclusive with P(A) = 0.4 and P(B) = 0.5. Are they independent? Explain.
No. P(A ∩ B) = 0, but P(A) × P(B) = 0.2. If A happens, B cannot.
3
A and B are independent, P(A) = 0.6 and P(B) = 0.25. Find P(A′ ∩ B′) and P(A′ ∩ B).
