الأسبوع 14 · الأحداث المستقلة وغير المستقلة: اختبار P(A ∩ B) = P(A) × P(B) والسحب مع الإرجاع وبدونه والاستقلال مقابل التنافي · 3 دروس
2
Using independence: both, at least one, only one and neither, with replacement
Statistics Sec 3 Term 1, Unit 3, Lesson 3
GoalUse independence to find P(both), P(at least one), P(only one) and P(neither), including draws with replacement.
New words
with replacement · the item is put back, so the draws are independentonly one · exactly one of the two events occursat least one · 1 − P(none), the fastest way
Explanation
If A and B are independent, the four regions of the Venn diagram follow from the two numbers P(A) and P(B). P(both) = P(A)P(B); P(A only) = P(A)(1 − P(B)); P(B only) = (1 − P(A))P(B); P(neither) = (1 − P(A))(1 − P(B)). Then “at least one” is 1 minus “neither”, and “exactly one” is A only plus B only. Two players take one free throw each, independently, with chances 0.6 and 0.7. Both score: 0.6 × 0.7 = 0.42. Only the first: 0.6 × 0.3 = 0.18. Only the second: 0.4 × 0.7 = 0.28. Neither: 0.4 × 0.3 = 0.12. The four add up to 1. For several tries the shortcut is P(at least one) = 1 − P(none). Drawing with replacement keeps draws independent, because the bag is the same every time.
Worked examples
Amr is on time with probability 0.8 and Salma with probability 0.9, independently. Both on time: 0.8 × 0.9 = 0.72. Neither: 0.2 × 0.1 = 0.02. At least one: 1 − 0.02 = 0.98. Exactly one: 0.8 × 0.1 + 0.2 × 0.9 = 0.08 + 0.18 = 0.26. Check: 0.72 + 0.26 + 0.02 = 1.
A bag holds 5 red and 2 blue counters. Two are drawn with replacement. P(RR) = 5/7 × 5/7 = 25/49; P(BB) = 2/7 × 2/7 = 4/49; P(R then B) = 5/7 × 2/7 = 10/49, and the same for B then R. One of each colour: 20/49. Check: 25 + 4 + 20 = 49.
Many tries. A die is thrown three times. P(no six) = (5/6)³ = 125/216, so P(at least one six) = 1 − 125/216 = 91/216, about 0.42. Counting the many ways to get a six directly would be much longer.
Class activity
Throw a die twice, 30 times, and count how often at least one six appears. Compare with 1 − 25/36 = 11/36 ≈ 0.31. Why does your count not have to be exactly 9 or 10?
Practice
1
Two buses are independently on time with probabilities 0.7 and 0.9. Find P(both), P(neither), P(exactly one) and P(at least one).
Both 0.63; neither 0.3 × 0.1 = 0.03; exactly one 0.7 × 0.1 + 0.3 × 0.9 = 0.34; at least one 0.97.
2
A and B are independent, P(A) = 0.5 and P(B) = 0.2. Find P(A only) and P(A ∪ B).
