Question

Statistic and Probability Question:

Three times each day a quality control engineer samples a component from a recently manufactured batch and tests it. Each part is given one of three classifications conforming - suitable for its intended use downgraded - unsuitable for its intended purpose, but usable for another purpose scrap - not usable An experiment consists of recording (in order) the classifications of the three parts tested on a particular day 1) List all of the outcomes in the sample space. (For example, two possible outcomes are CCS and DSC. CCS means that the first two parts were conforming, and the third was scrap. DSC means that the first was downgraded, the second was scrap, and the third was conforming. Note that this is not the same as CSD.) How many are there? 27 2) Let A be the event that all the parts fall into the same category. How many outcomes are in event A? 3 3) Let B be the event that at least two parts are conforming. How many outcomes are in event B? 6 4 How many outcomes are in the event AnB? b) 4 d) 3 5) How many outcomes are in the eventAnB' b) 4 d) 1

** I ONLY NEED 4 AND 5.

Answer 1

Answer

(4) Possible outcomes for event A = {CCC, DDD, SSS}

because we need all the parts fall into the same category

So, there are three possible outcomes for event A

Possible outcomes for event B = {CCD, CCS, CSC, CDC, DCC, SCC, CCC}

because we need at least two conforming parts.

So, there are three possible outcomes for event B

Now, we have to find the intersection of A and B, i.e. common or same element in both sets

we can see that only element CCC is common or same in both events A and B

So, we can write $A \cap B = {CCC}$

So, there is only 1 common element

Option A is correct answer

(5)Possible outcomes for event A = {CCC, DDD, SSS}

because we need all the parts fall into the same category

So, there are three possible outcomes for event A

Possible outcomes for event B = {CCD, CCS, CSC, CDC, DCC, SCC, CCC}

because we need at least two conforming parts.

So, there are three possible outcomes for event B

Now, we have to find the intersection of A and B', i.e. common or same element in A and B'

we know that B' or B complement includes all outcomes except the outcomes included in event B

i.e. B' will contain all elements except the elements of set {CCD, CCS, CSC, CDC, DCC, SCC, CCC}

So, B' will have 27-7 = 20 elements

Event A include only 3 elements. Out of 3 elements of A, one element (CCC) is included in B, which we will not count in B'

So, we have only 2 common outcomes or elements in intersection of A and B', i.e. $A \cap B' = (DDD, SSS)$

So, number of common outcomes is 2

option A is correct answer