Answer to Question #244672 in Statistics and Probability for Nana

Question

In a production process, a total number of 100 items were produced by machines M1, M2 and M3. Of those items being produced, 37 were made by machine M1, 42 were made by M2, and 21 were made by M3. The three machines work independently, however, they do not work perfectly. From the experience, 5% of the items produced by M1 are defective, 4% of the items produced by M2 are defective, and 3% of the items produced by M3 are defective.

a) Given that a randomly selected item is non-defective, what is the probability that it is produced by machine M1?

(b) If two items which are not produced by machine M3 are selected at random without replacement, what is the probability that at least one of them is non-defective? 

Answer 1

Probabilities for production of items for the three machines are given as,

$p(M1)=37/100=0.37$

$p(M2)=42/100=0.42$

$p(M3)=21/100=0.21$

Let D denote defective and N be non-defective items produced by the machines.

Number of defective and non-defective items per machine are as follows,

5% of items produced by M1 are defective therefore, number of defective items for M1 are,

D=5/100*37=1.85$\approxeq2$

N=37-2=35

4% of items produced by M2 are defective therefore, number of defective items for M2 are,

D=4/100*42=1.68$\approxeq2$

N=42-2=40

3% of items produced by M3 are defective thus, number of defective items for M3 are,

D=3/100*21=0.63$\approxeq1$

N=21-1=20

Conditional probabilities of non-defective items for each machine are given as,

$p(N|M1)=35/37$

$p(N|M2)=40/42$

$p(N|M3)=20/21$

Probability that a non-defective item is randomly selected is determined using the law of total probability illustrated below.

$p(N)=p(N|M1)*p(M1)+p(N|M2)*p(M2)+p(N|M3)*p(M3)$

$p(N)=(35/37)*0.37+(40/42)*0.42+(20/21)*0.21=0.95$

(a).

Given that a non-defective item is randomly selected, the probability that it is produced by M1 is given as,

$p(M1|N)=p(N|M1)*p(M1)/p(N)$

$p(M1|N)=(35/37*0.37)/0.95=0.3684(4\displaystyle\space decimal\displaystyle\space places)$

(b).

Machines M1 and M2 are considered in this case.

$p(N\geqslant1|M1)=1190/1332+70/1332+70/1332=1330/1332=0.9984985$

$p(N\geqslant1|M2)=1560/1722+80/1722+80/1722=1720/1722=0.99883856$

$p(N\geqslant1)=p(M1)*p(N\geqslant1|M1)+p(M2)*p(N\geqslant1|M2)$

$p(N\geqslant1)=(0.37*0.9984985)+(0.42*0.99883856)=0.7889566$

Therefore, probability that at least one of them is non-defective is 0.789(3 decimal places).