Example 2. An electronic company manufactures computers,...

Example 2. An electronic company manufactures computers, televisions, CD players, and others electronics products. When shipped items are damaged, the causes of the damage are categorized as water (W), crushing (C), puncture (P), or carton marking (M). Listed below are the coded causes of five damaged items. A quality control analyst wants to randomly select two items for further investigation Find the probability that the first selected item was damaged from crushing (C) and the second item was also damaged crushing (C). Assume that the selections are made (a) with replacement; (b) without replacement. W C C P M

Transcript

0:00 Hello everyone.

00:02 So in this question, there are three factories 1, 2 and 3.

00:07 Let us first define our events.

00:13 This is for part 8.

00:16 This is for part 8.

00:18 So events are f1, f2, f3 which is produced in factory 1.

00:37 Similarly produced in factory 2 and third is produced in factory 3 also let us define even d which is defective now as per the given information let number of items of items produced in factory 2 b x then aspati given information number of items produced in factory 3 becomes same as those produced in 2 that is x whereas number of items produced in factory 1 is twice that produced in 2 and 3 that is 2x thus we get probability of f1 is equal to 2x divided by 4x which is 1 by 2 probability of f2 is x divided by 4x 1 by 4 x divided by 4 and probability of f3 is x divided by 4x also now here we have to find the probability of d that is probability of defective which is same as probability of f1 multiplied by probability of defective given f1 plus probability reduced in factory 2 multiplied by probability it is defective given produced in factory 2 plus probability produced in factory 3 multiplied by probability it is defective given produced in factory 3.

03:36 Substituting the values be yet 1 divided by 2 multiplied by, as per the given information, it is 2 % that is 2 divided by 100 plus 1 divided by 4 multiplied by 1 .000, again 2 % 2 divided by 100 plus 1 divided by 4 multiplied by 4%, that is 4 divided by 100.

03:59 On solving, we give this to be equals to 0 .025, that is 2 .5%.

04:09 This is the solution for part 8.

04:11 Now, let us come to part b.

04:16 Part b.

04:18 Let us again take the events, say a and b this time.

04:26 A being that chip is stolen and b being that chip is defective.

04:42 Now, as but the given information, the probabilities are given to us here is probability of a...

Question

Please give Ace some feedback