Question 2 OR (d)

Bayes' theorem: who was selected, given that the structure changed

6 marks

Time to write: About 12 minutes
Topic: Probability and Bayes' theorem

Question

Question 2 OR (d) · 6 marks

A, B, and C are candidates for the post of director at a private firm, and their chances of selection are in the ratio 4:3:3. The probabilities that they will make changes in the organization structure are 0.4, 0.7, and 0.6, respectively. Given that a change has occurred, find the probability that:

It resulted from B's selection,
It resulted from A's selection.

Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.

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Read off the question and its marks

12 minfor 6 marks

Read from the question

findFormula first, then the working, then the answer with its unit (₹, %, units).

Write it in this order

Read the data; note exactly what is asked 1 min
1 Events, priors and conditional probabilities 1½ marks 3 min
2 Joint probabilities and P(E) 1½ marks 3 min
3 P(B \ 1½ marks 2 min
4 P(A \ 1½ marks 2 min
Check: arithmetic, units (₹, %), final answer underlined 1 min

1234

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Solution

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Events. Let A, B and C be the events that A, B or C is selected, and E the event that the organisation structure is changed. A, B and C are mutually exclusive and exhaustive (exactly one director is chosen), with P(A): P(B): P(C) = 4: 3: 3, so P(A) = 4/10 = 0.4, P(B) = 0.3 and P(C) = 0.3. The conditional probabilities of a change are P(E | A) = 0.4, P(E | B) = 0.7 and P(E | C) = 0.6.

By the multiplication theorem, P(A ∩ E) = P(A) × P(E | A), and so on. Adding the three mutually exclusive joint probabilities gives P(E) (total probability), and by the definition of conditional probability, P(A | E) = P(A ∩ E) / P(E). Together these give Bayes' theorem:

P(Aᵢ | E) = P(Aᵢ) P(E | Aᵢ) / Σ P(Aⱼ) P(E | Aⱼ).

Working table

Candidate	Prior P(Aᵢ)	P(E \| Aᵢ)	Joint P(Aᵢ) × P(E \| Aᵢ)	Posterior P(Aᵢ \| E)
A	0.4	0.4	0.16	16/55 = 0.2909
B	0.3	0.7	0.21	21/55 = 0.3818
C	0.3	0.6	0.18	18/55 = 0.3273
Total	1		0.55	1

So P(E) = 0.16 + 0.21 + 0.18 = 0.55.

(i) Probability that the change came from B's selection

P(B | E) = 0.21 / 0.55 = 21/55 = 0.3818.

(ii) Probability that the change came from A's selection

P(A | E) = 0.16 / 0.55 = 16/55 = 0.2909.

Comment. Before anything is known, A is the likeliest choice (0.4). Once a change has occurred, B becomes the likeliest (0.3818), because B is far more likely than A to change the structure (0.7 against 0.4).

Where the 6 marks sit

#	Component	Marks
1	Events, priors and conditional probabilities	1.5
2	Joint probabilities and P(E)	1.5
3	P(B \| E) by Bayes' theorem	1.5
4	P(A \| E) by Bayes' theorem	1.5
	Total	6

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