If 5 of 20 tires in storage are defective and 5 of them are randomly chosen for inspection
(that is, each tire has the same chance of being selected), what is the probability that
the two of the defective tires will be included?
Answers
Given : 5 of 20 tires in storage are defective
and 5 of them are randomly chosen for inspection
To Find : probability that the two of the defective tires will be included?
Solution:
Case 1: There are 20 tyres only
Defective = 5
Non defective = 20 - 5 = 15 tyres
5 of them are randomly chosen for inspection
Ways = ²⁰C₅ = 15,504
two of the defective tires will be included
= ⁵C₂.¹⁵C₃
= 10 . 455
= 4550
probability that the two of the defective tires will be included = 4550/15,504
= 0.2935
Case 2 : There are lot of tires on average 5 of 20 tires in storage are defective
Defective probability = 5/20 = 1/4
q = 1 - p = 3/4
n = 5
x = 2
P(x) = ⁿCₓpˣqⁿ⁻ˣ
=> P(2) = ⁵C₂(1/4)²(3/4)³ = 10 * 27 / 4⁵ = 270/1024 = 0.2637
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