Warehouse Order-Fulfilment Tree Diagram: 04-BS-2, May 2014, Q4

Question 4: Warehouse Order-Fulfilment Tree Diagram (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. P(A)=0.30, P(B)=0.25, P(C)=0.20, P(D)=0.25 (order-filling shares); conditional mistake rates P(M^C|A)=0.010, P(M^C|B)=0.012, P(M^C|C)=0.014, P(M^C|D)=0.011.

Find. (a) the tree diagram; (b) Pr(M), Pr(B∩M), Pr(D∩M^C); (c) Pr(C | mistake); (d) P(at least 14 of 15 orders correctly filled).

Approach. Build the two-stage tree (employee, then mistake/no-mistake), apply the multiplication rule along each branch, then the law of total probability for Pr(M) and Bayes' theorem for the reversed conditional in (c). Part (d) treats each order as an independent Bernoulli trial with success probability Pr(M) found in (b)(i).

(a) Tree diagram. Each first-stage branch (A, B, C, D) carries the employee's order share; each second-stage branch (M, M^C) carries that employee's conditional mistake/no-mistake probability.
Tree diagram: first-stage branches A/B/C/D (order-filling probabilities), second-stage branches M (no mistake) / Mᶜ (mistake) with their conditional probabilities.
(b)(i) Pr(M) by the law of total probability. $$\Pr(M^C)=\Pr(A)P(M^C|A)+\Pr(B)P(M^C|B)+\Pr(C)P(M^C|C)+\Pr(D)P(M^C|D)$$ $$=0.30(0.010)+0.25(0.012)+0.20(0.014)+0.25(0.011)=0.01155,$$ $$\Pr(M)=1-\Pr(M^C)=1-0.01155=\boxed{0.98845}.$$ (b)(ii) Along the B-branch, $\Pr(B\cap M)=\Pr(B)\big[1-P(M^C|B)\big]=0.25(0.988)=\boxed{0.2470}.$ (b)(iii) Along the D-branch, $\Pr(D\cap M^C)=\Pr(D)\,P(M^C|D)=0.25(0.011)=\boxed{0.00275}.$
(c) Bayes' theorem for Pr(C | mistake). Given a wrong item was shipped (event M^C), the probability employee C caused it is $$\Pr(C\mid M^C)=\dfrac{\Pr(C)P(M^C|C)}{\Pr(M^C)}=\dfrac{0.20(0.014)}{0.01155}=\dfrac{0.0028}{0.01155}=\boxed{0.2424}.$$
(d) At least 14 of 15 orders correctly filled. Treating each of the 15 orders as an independent Bernoulli trial with $p=\Pr(M)=0.98845$ (from part (b)(i)), let $W\sim\text{Binomial}(15,\,0.98845)$ count correctly-filled orders. Then $$P(W\ge14)=P(14)+P(15)=15(0.98845)^{14}(0.01155)+(0.98845)^{15}=0.1442+0.8432=\boxed{0.9873}.$$

Final results — Question 4
Part	Quantity	Result
(b)(i)	Pr(M)	0.98845
(b)(ii)	Pr(B∩M)	0.2470
(b)(iii)	Pr(D∩M^C)	0.00275
(c)	Pr(C \| mistake)	0.2424
(d)	P(at least 14 of 15 correct)	0.9873

Topic: The multiplication rule, law of total probability, Bayes' theorem, and their translation into a tree diagram.

Key relations: $\Pr(E_i\cap F)=\Pr(E_i)P(F|E_i)$; $\Pr(F)=\sum_i \Pr(E_i)P(F|E_i)$; $\Pr(E_i|F)=\dfrac{\Pr(E_i)P(F|E_i)}{\Pr(F)}$.

Why this works: the tree's branch probabilities are joint probabilities by construction (multiplication rule along a path); summing the M^C-branch endpoints recovers the marginal Pr(M^C) because the four employee events partition the sample space; Bayes' theorem then just re-expresses one joint probability (C∩M^C) as a fraction of that same marginal.

Common pitfall: computing Pr(C|mistake) as P(M^C|C)=0.014 directly — that is the reverse conditional (probability C errs, given C filled the order), not the probability that C is the culprit given that some mistake occurred; the denominator must be the marginal Pr(M^C), not 1.

Source: Walpole, Myers, Myers & Ye, Ch. 2 (Probability — Bayes' Theorem).

Question 4 of 8: Warehouse Order-Fulfilment Tree Diagram

Question 4: Warehouse Order-Fulfilment Tree Diagram (20 marks)