CBSE Sample Paper 2020-21 · Part A: Section II · 4 marks

Q18: Case study: three employees, their error rates and Bayes' theorem

Sample 2020-21Ch 13 · Probability

The question

In an office three employees Vinay, Sonia and Iqbal process incoming copies of a certain form. Vinay process 50% of the forms. Sonia processes 20% and Iqbal the remaining 30% of the forms. Vinay has an error rate of 0.06, Sonia has an error rate of 0.04 and Iqbal has an error rate of 0.03

Based on the above information answer the following:

(i) The conditional probability that an error is committed in processing given that Sonia processed the form is : (1 mark)

a) 0.0210

b) 0.04

c) 0.47

d) 0.06

(ii) The probability that Sonia processed the form and committed an error is : (1 mark)

a) 0.005

b) 0.006

c) 0.008

d) 0.68

(iii) The total probability of committing an error in processing the form is (1 mark)

a) 0

b) 0.047

c) 0.234

d) 1

(iv) The manager of the company wants to do a quality check. During inspection he selects a form at random from the days output of processed forms. If the form selected at random has an error, the probability that the form is NOT processed by Vinay is : (1 mark)

a) 1

b) 30/47

c) 20/47

d) 17/47

(v) Let A be the event of committing an error in processing the form and let $E_1$ ⁠, $E_2$ and $E_3$ be the events that Vinay, Sonia and Iqbal processed the form. The value of $\sum_{i=1}^{3} P(E_i \mid A)$ is (1 mark)

a) 0

b) 0.03

c) 0.06

d) 1

Exam answer

Any 4 of the 5 sub parts are to be attempted; all 5 are answered here.

Let $E_1$ ⁠, $E_2$ ⁠, $E_3$ be the events that Vinay, Sonia and Iqbal processed the form, and A the event that an error is committed.

\begin{aligned} &P(E_1) = 0.5,\ P(E_2) = 0.2, \\ &P(E_3) = 0.3 \end{aligned}

\begin{aligned} &P(A|E_1) = 0.06,\ P(A|E_2) = 0.04, \\ &P(A|E_3) = 0.03 \end{aligned}

Why?

The shares of the forms are probabilities: 50% = 0.5, 20% = 0.2 and 30% = 0.3. They add to 1, because every form is processed by exactly one of the three.

An error rate applies once we know who processed the form, so it is a conditional probability. Vinay's rate 0.06 is $P(A|E_1)$ ⁠, not $P(A)$ ⁠.

(i) Given that Sonia processed the form, the probability of an error is Sonia's error rate:

Why?

"Given that Sonia processed the form" is the condition $E_2$ ⁠. So the question asks for $P(A|E_2)$ ⁠, which is stated in the case study.

P(A|E_2) = 0.04

Why?

No calculation is needed: the error rate of a person is already a conditional probability.

Watch out: this is not $P(E_2 \cap A)$ ⁠. That is part (ii).

Answer: (b) 0.04

(ii) Sonia processed the form and committed an error: this is the event $E_2 \cap A$ ⁠.

Why?

"And" means both events happen, so we need the intersection $E_2 \cap A$ ⁠, a joint probability.

\begin{aligned} P(E_2 \cap A) &= P(E_2)\,P(A|E_2) \\ &= 0.2 \times 0.04 = 0.008 \end{aligned}

Why?

The multiplication rule gives

P(E \cap A) = P(E)\,P(A|E)

Arithmetic: $2 \times 4 = 8$ ⁠, and the decimal places add up: $1 + 2 = 3$ ⁠.

Answer: (c) 0.008

(iii) By the theorem of total probability,

Why?

Every error happens on a form processed by Vinay, Sonia or Iqbal, and these three cases do not overlap. So $P(A)$ is the sum of $P(E_i \cap A)$ over the three, and each term comes from the multiplication rule of (ii).

\begin{aligned} P(A) &= \sum_{i=1}^{3} P(E_i)\,P(A|E_i) \\ &= 0.5 \times 0.06 + 0.2 \times 0.04 \\ &\quad + 0.3 \times 0.03 \\ &= 0.030 + 0.008 + 0.009 \\ &= 0.047 \end{aligned}

Why?

The three products are $0.5 \times 0.06 = 0.030$ ⁠, $0.2 \times 0.04 = 0.008$ and $0.3 \times 0.03 = 0.009$ ⁠.

Multiply the digits, then count the decimal places ( $1 + 2 = 3$ ⁠):

\begin{aligned} 0.5 \times 0.06 &: 5 \times 6 = 30 \to 0.030 \\ 0.2 \times 0.04 &: 2 \times 4 = 8 \to 0.008 \\ 0.3 \times 0.03 &: 3 \times 3 = 9 \to 0.009 \end{aligned}

Then add two at a time:

\begin{aligned} 0.030 + 0.008 &= 0.038 \\ 0.038 + 0.009 &= 0.047 \end{aligned}

Watch out: adding the error rates alone, $0.06 + 0.04 + 0.03 = 0.13$ ⁠, is wrong. Each rate must be weighted by that person's share of the forms.

Answer: (b) 0.047

(iv) By Bayes' theorem,

\begin{aligned} P(E_1|A) &= \frac{P(E_1)\,P(A|E_1)}{P(A)} \\ &= \frac{0.030}{0.047} = \frac{30}{47} \end{aligned}

Why?

Now we know the result (an error) and ask who caused it. Reversing the condition like this is Bayes' theorem: the part of $P(A)$ that comes from Vinay, divided by the whole $P(A)$ ⁠.

The top $0.030$ was found in (iii). Multiplying the top and bottom by 1000 gives $\frac{30}{47}$ ⁠.

\begin{aligned} \frac{0.030}{0.047} &= \frac{0.030 \times 1000}{0.047 \times 1000} \\ &= \frac{30}{47} \end{aligned}

\begin{aligned} P(\bar E_1|A) &= 1 - P(E_1|A) \\ &= 1 - \frac{30}{47} = \frac{17}{47} \end{aligned}

Why?

"Not processed by Vinay" is the complement of $E_1$ ⁠. Given the error, the two cases add to 1.

\begin{aligned} 1 - \frac{30}{47} &= \frac{47}{47} - \frac{30}{47} \\ &= \frac{47 - 30}{47} = \frac{17}{47} \end{aligned}

Check: Sonia and Iqbal together give $\frac{0.008 + 0.009}{0.047} = \frac{17}{47}$ ⁠.

\begin{aligned} \frac{0.008 + 0.009}{0.047} &= \frac{0.017}{0.047} \\ &= \frac{17}{47} \quad (\times 1000) \end{aligned}

Answer: (d) 17/47

(v) By Bayes' theorem, each $P(E_i|A) = \frac{P(E_i)\,P(A|E_i)}{P(A)}$ ⁠. So

Why?

This is the same formula as in (iv), used for each of the three people. All three fractions have the same denominator $P(A)$ ⁠, so they add by adding the numerators.

\begin{aligned} &\sum_{i=1}^{3} P(E_i|A) \\ &= \frac{\sum_{i=1}^{3} P(E_i)\,P(A|E_i)}{P(A)} \\ &= \frac{P(A)}{P(A)} = 1 \end{aligned}

Why?

By (iii), the numerators add to $P(A)$ ⁠:

\begin{aligned} &\frac{0.030 + 0.008 + 0.009}{0.047} \\ &= \frac{0.047}{0.047} = 1 \end{aligned}

In words: given an error, the form was certainly processed by one of the three people.

Watch out: the sum of the error rates $P(A|E_i)$ is $0.13$ ⁠, not 1. The sum that equals 1 has $A$ as the condition.

Answer: (d) 1

Where this comes from

Chapter 13, Probability: Total probability and Bayes' theorem.

Question type: (ii) One after the other; (iii) Two routes to one result; (iv) Bayes' theorem; (v) Bayes' theorem

How to spot it: click to see

(ii) Type 3. One after the other: the multiplication rule

Spot it: draws "without replacement", or a second event that depends on the first.

Here: "Sonia processed the form and committed an error" is one thing after another: Sonia first, $P(E_2) = 0.2$ ⁠, then an error given Sonia, 0.04. Multiply: 0.008.

(iii) Type 7. Two routes to one result: total probability

Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.

Here: An error can come from Vinay, Sonia or Iqbal. Multiply each share by that person's error rate and add: $0.030 + 0.008 + 0.009 = 0.047$ ⁠.

(iv) Type 8. Bayes' theorem: from the result back to the cause

Spot it: the result has already happened (a red ball, a faulty item, an A grade), and you are asked for the chance that it came from one cause. It often comes as a case study or a 5-mark question.

Here: The form already has an error, and you want the chance of its cause. Vinay's route is $\frac{0.030}{0.047} = \frac{30}{47}$ ⁠, so not Vinay is $1 - \frac{30}{47}$ ⁠.

(v) Type 8. Bayes' theorem: from the result back to the cause

Here: Each $P(E_i \mid A)$ is one route over the total $P(A)$ ⁠. Adding all three puts $P(A)$ over $P(A)$ ⁠, so the sum is 1.

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