CBSE Sample Paper 2023-24 · Section E · 4 marks

Q36: Case study: error rates of three employees processing forms

Sample 2023-24Ch 13 · Probability

The question

Read the following passage and answer the questions given below:

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

Based on the above information, answer the following questions.

(i) Find the probability that Sonia processed the form and committed an error. (1 mark)

(ii) Find the total probability of committing an error in processing the form. (1 mark)

(iii) (a) 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 form. If the form selected at random has an error, find the probability that the form is not processed by Jayant. (2 marks)

(iii) (b) Let $E$ be the event of committing an error in processing the form and let $E_1$ ⁠, $E_2$ and $E_3$ be the events that Jayant, Sonia and Oliver processed the form. Find the value of $\displaystyle\sum_{i=1}^{3} P(E_i \mid E)$ ⁠. (2 marks)

Exam answer

Let $E_1$ ⁠, $E_2$ ⁠, $E_3$ be the events that Jayant, Sonia and Oliver processed the form, which are pairwise mutually exclusive and exhaustive. Then

\begin{aligned} P(E_1) &= \frac{50}{100} = \frac{5}{10} \\ P(E_2) &= \frac{20}{100} = \frac15 \\ P(E_3) &= \frac{30}{100} = \frac{3}{10} \end{aligned}

Why?

Each form is processed by exactly one of the three, so the events do not overlap and together they cover every form. A percentage becomes a probability when divided by 100. Check: $0.5 + 0.2 + 0.3 = 1$ ⁠.

Also, let $E$ be the event of committing an error.

\begin{aligned} P(E \mid E_1) &= 0.06 \\ P(E \mid E_2) &= 0.04 \\ P(E \mid E_3) &= 0.03 \end{aligned}

Why?

An error rate is a conditional probability. "Jayant has an error rate of 0.06" means: given that Jayant processed the form, the chance of an error is 0.06.

(i) The probability that Sonia processed the form and committed an error is

\begin{aligned} P(E \cap E_2) &= P(E_2) \cdot P(E \mid E_2) \\ &= \frac15 \times 0.04 = 0.008 \end{aligned}

Why?

"Sonia processed it and there is an error" is the event $E \cap E_2$ ⁠. By the multiplication rule,

P(A \cap B) = P(B)\,P(A \mid B)

$\frac15 = 0.2$ ⁠, and $0.2 \times 0.04 = 0.008$ ⁠.

\begin{aligned} \frac15 &= \frac{1 \times 2}{5 \times 2} = \frac{2}{10} = 0.2 \\ 0.2 \times 0.04 &= \frac{2}{10} \times \frac{4}{100} \\ &= \frac{8}{1000} = 0.008 \end{aligned}

Watch out: 0.04 alone is the chance of an error once we know Sonia did it. It is not the chance of both happening.

The required probability is 0.008.

(ii) The total probability of committing an error is

\begin{aligned} P(E) &= P(E_1) \cdot P(E \mid E_1) \\ &\quad + P(E_2) \cdot P(E \mid E_2) \\ &\quad + P(E_3) \cdot P(E \mid E_3) \\ &= \frac{50}{100} \times 0.06 + \frac{20}{100} \times 0.04 \\ &\quad + \frac{30}{100} \times 0.03 \\ &= 0.03 + 0.008 + 0.009 \\ &= 0.047 \end{aligned}

Why?

An error can come through Jayant, through Sonia or through Oliver, and these routes do not overlap. The theorem of total probability adds the three routes, each weighted by how often that person works:

P(E) = \sum_{i=1}^{3} P(E_i)P(E \mid E_i)

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

\begin{aligned} 0.5 \times 0.06 &= 0.030 \quad (5 \times 6 = 30) \\ 0.2 \times 0.04 &= 0.008 \quad (2 \times 4 = 8) \\ 0.3 \times 0.03 &= 0.009 \quad (3 \times 3 = 9) \\[4pt] 0.030 + 0.008 &= 0.038 \\ 0.038 + 0.009 &= 0.047 \end{aligned}

Each product has three decimal places, because the two factors have one and two.

The total probability of committing an error is 0.047.

(iii) (a) By Bayes' theorem, the probability that the form is processed by Jayant, given that it has an error, is

\begin{aligned} P(E_1 \mid E) &= \frac{P(E_1) \cdot P(E \mid E_1)}{P(E)} \\ &= \frac{0.06 \times \frac{50}{100}}{0.047} = \frac{30}{47} \end{aligned}

Why?

We know the result (an error) and ask who caused it. Reversing a conditional probability is the job of Bayes' theorem: the share of the error probability that comes through Jayant.

The top is $0.06 \times 0.5 = 0.03$ (from part (ii)). Multiply top and bottom by 1000: $\frac{0.03}{0.047} = \frac{30}{47}$ ⁠.

\begin{aligned} \frac{0.03}{0.047} &= \frac{0.03 \times 1000}{0.047 \times 1000} \\ &= \frac{30}{47} \end{aligned}

Therefore, the required probability that the form is not processed by Jayant, given that it has an error, is

\begin{aligned} P(\overline{E_1} \mid E) &= 1 - P(E_1 \mid E) \\ &= 1 - \frac{30}{47} = \frac{17}{47} \end{aligned}

Why?

Given an error, the form was processed either by Jayant or not by Jayant, so the two conditional probabilities add to 1.

Write 1 as $\frac{47}{47}$ to subtract:

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

Check another way, Sonia's and Oliver's shares: $\frac{0.008 + 0.009}{0.047} = \frac{17}{47}$ ⁠.

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

The required probability is $\frac{17}{47}$ ⁠.

(iii) (b)

\begin{aligned} &\sum_{i=1}^{3} P(E_i \mid E) \\ &= P(E_1 \mid E) + P(E_2 \mid E) \\ &\quad + P(E_3 \mid E) \\ &= \frac{1}{P(E)}\bigg[P(E \cap E_1) + P(E \cap E_2) \\ &\quad + P(E \cap E_3)\bigg] \\ &= \frac{P(E \cap (E_1 \cup E_2 \cup E_3))}{P(E)} \\ &= \frac{P(E \cap S)}{P(E)} = \frac{P(E)}{P(E)} = 1 \end{aligned}

Why?

By the definition of conditional probability, each term is

P(E_i \mid E) = \frac{P(E \cap E_i)}{P(E)}

All three have the same bottom $P(E)$ ⁠, so it is taken out. The three pieces $E \cap E_i$ do not overlap, so their probabilities add to the probability of their union, which is all of $E$ ⁠.

Here $E \cap E_i$ and $E \cap E_j$ ( $i \ne j$ ⁠) are mutually exclusive, and $E_1 \cup E_2 \cup E_3 = S$ ⁠, the sample space.

Why?

A numerical check with the values found above. Each $P(E_i \mid E)$ is that route over $P(E) = 0.047$ ⁠:

\begin{aligned} P(E_1 \mid E) &= \frac{0.030}{0.047} = \frac{30}{47} \\ P(E_2 \mid E) &= \frac{0.008}{0.047} = \frac{8}{47} \\ P(E_3 \mid E) &= \frac{0.009}{0.047} = \frac{9}{47} \end{aligned}

Add the tops: $30 + 8 + 9 = 47$ ⁠.

\frac{30}{47} + \frac{8}{47} + \frac{9}{47} = \frac{47}{47} = 1

$\therefore\ \displaystyle\sum_{i=1}^{3} P(E_i \mid E) = 1$ (the sum of the posterior probabilities is 1).

Where this comes from

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

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

How to spot it: click to see

(i) 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, $P(E_2) = \frac15$ ⁠, then an error given Sonia, 0.04. Multiply: 0.008.

(ii) 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 Jayant, Sonia or Oliver. Multiply each share by that person's error rate and add: $0.03 + 0.008 + 0.009 = 0.047$ ⁠.

(iii)(a) 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. Jayant's route gives $\frac{30}{47}$ ⁠, so not Jayant is $1 - \frac{30}{47} = \frac{17}{47}$ ⁠.

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

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

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