Question

Problem no. 20 An engineer responsible for the control of the quality in the company where she works receives a batch of 200 parts used in the computers that they build. She decides to pick 10, at random and without replacement, and to test them. Let E be the random experiment that consists in counting the number of defective parts among 2.6 Exercises, Problems, and Multiple Choice Questions 37 the I0 parts tested. Answer the following questions by assuming that there are in fact two defective parts among the 200 parts received a) Write the sample space S. b) Calculate the probability of all the elementary events. c) We define the event Ak - there are exactly k defective parts among the 10 parts tested, fork 0, I, . . . . Calculate PIAoU (Ain A2)].

Answer 1

a)

Since, there are only two defective parts among the 200 parts received, maximum number of defective parts among 10 parts selected without replacement is 2.

So, the sample space for number of defective parts in the experiment E is S = {0, 1, 2}

b)

Number of non-defective parts = 200 - 2 = 198

Number of ways to select 10 parts from 200 = ²⁰⁰C₁₀

Number of ways to select k defective parts from 2 = ²C_k   (where k = 0, 1, 2)

Number of ways to select k non-defective parts from 198 = ¹⁹⁸C_k   (where 0 $\le$ k $\le$ 198)

Probability of selecting 0 defectives in a sample of 10 parts = Probability of selecting 0 defectives from 2 defective and 10 non-defectives from 198 non-defectives parts

= (²C₀ * ¹⁹⁸C₁₀)/ ²⁰⁰C₁₀ = 0.9022613

Probability of selecting 1 defectives in a sample of 10 parts = Probability of selecting 1 defectives from 2 defective and 9 non-defectives from 198 non-defectives parts

= (²C₁ * ¹⁹⁸C₉)/ ²⁰⁰C₁₀ = 0.09547739

Probability of selecting 2 defectives in a sample of 10 parts = Probability of selecting 2 defectives from 2 defective and 8 non-defectives from 198 non-defectives parts

= (²C₀ * ¹⁹⁸C₁₀)/ ²⁰⁰C₁₀ = 0.002261307

c)

As k cannot take values 0, 1 or 2 at the same time, A0 , A1 and A2 are disjoint events and thus,

$A_1^c \cap A_2 = A_2$

So,

$P[A_0 \cup (A_1^c \cap A_2)] = P[A_0 \cup A_2] = P[A_0] + P[A_2]$ (A0 and A2 are disjoint events)

= 0.9022613 + 0.002261307

= 0.9045226