Question

Please answer the questions clearly

11. Data shows that approximately 3% of people have a specific disease. There is a test for this (liscase, bui ihe rot;ulį, IS 110l alway:; accurie. [1 a :;lek person i aks: í he io:;i, i here is 98% chanc that the result of the test will be positive. However, if a healthy person takes the test, there is 97% chance that the result of the test will be negative. If a random person from the population takes the test, (a) What is the probability that the result will be negative? b) Given the result is negative, what is the probability that the person who took the test is actually sick? 12. One division of an Amazon fulfillment center employs 3 people, U, V, and W, who pul items from shelves and assemble them for subsequent verification and packaging. U makes a mistake in an order (gets a wrong item or the wrong quantity) one time in a hundred, V makes a mistake in an order five times in a hundred, and W makes a mistake in an order three times in a hundred. If U, V, and W fill, respectively, 30, 40, and 30 percent of all orders, what is the probability that if a mistake is made in an order, the order was filled by U?

Answer 1

11.

Let the event that the test result is negative be denoted as N $\Rightarrow$ The event that the test result is positive be denoted as N^c

Let the event that a random person is sick be denoted as S $\Rightarrow$ The event that a random person is not sick be denoted as S^c  

Given :- P(S)=0.03 $\Rightarrow$ P(S^c)=1-0.03=0.97

If a sick person takes the test, there is 98% chance that the test result is positive $\Rightarrow$ P(N^c|S)=0.98 $\Rightarrow$ P(N|S)=1-0.98=0.02

If a healthy person takes the test, there is 97% chance that the test result is negative $\Rightarrow$ P(N|S^c)=0.97

(a)

If a random person takes the test, probability that the test result will be negative = P(N)

According to the theorem of Total Probability :-

P(N) = P(N$\cap$S) + P(N$\cap$S^c) = P(N|S)$\times$P(S) + P(N|S^c)$\times$P(S^c) = 0.02$\times$0.03 + 0.97$\times$0.97 = 0.9415

(b)

Given the result is negative, probability that the person taking the test is actually sick = P(S|N)

P(S|N) = P(S$\cap$N)/P(S) = P(N$\cap$S)/P(S) = [P(N|S)$\times$P(S)]/P(N) = (0.02$\times$0.03)/0.9415 = 0.0006373

12.

Let the event that a mistake is made be denoted as M

Let the event that U makes a mistake be denoted as U_m

Let the event that V makes a mistake be denoted as V_m

Let the event that W makes a mistake be denoted as W_m

Let the event that U fills an order be denoted as U_f

Let the event that V fills an order be denoted as V_f

Let the event that W fills an order be denoted as W_f

According to the problem ; event U_m$\equiv$ event M|U_f event V_m$\equiv$ event M|V_f event W_m$\equiv$ event M|W_f

Given ; P(U_m)=1/100=0.01 ; P(V_m) =5/100=0.05 ; P(W_m)=3/100=0.03

P(U_f)=30%=0.3 ; P(V_f)=40%=0.4 ; P(W_f)=30%=0.3

According to the theorem of Total Probability :-

P(M) = P(M$\cap$U_f) + P(M$\cap$V_f) + P(M$\cap$W_f) = P(M|U_f)$\times$P(U_f) + P(M|V_f)$\times$P(V_f) + P(M|W_f)$\times$P(W_f)

= P(U_m)$\times$P(U_f) + P(V_m)$\times$P(V_f) + P(W_m)$\times$P(W_f)

= 0.01$\times$0.3 + 0.05$\times$0.4 + 0.03$\times$0.3 = 0.032

If a mistake is made in order, probability that the order was filled by U = P(U_f|M)

P(U_f|M) = P(U_f$\cap$M)/P(M) = P(M$\cap$U_f)/P(M) = [P(M|U_f)$\times$P(U_f)]/P(M) = [P(U_m)$\times$P(U_f)]/P(M) = (0.01$\times$0.3)/0.032 = 0.09375