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. (ii) Find the total probability of committing an error in processing the form. (iii) 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. (iii) 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 i=1³P(E_i E).

MathsCBSE Class 12 · Sample paper, session 2023-24 · 4 marks · Case-based

CBSE Class 12 Mathematics sample paper 2023-24: question 36

From the official CBSE sample question paper for 2023-24, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedProbability

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)
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
OR
(iii)
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 $\sum_{i = 1}^{3} P (E_{i} ∣ E)$ .
2 marks

Figure in the paper: Photograph of three office employees looking at papers and a laptop. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write the prior probabilities of each person and the conditional error probabilities.

Hint 2

Total probability: add $P (E_{i}) P (E ∣ E_{i})$ over all three; Bayes' theorem divides one such term by this total.

Hint 3

For 'not Jayant', use the complement of $P (E_{1} ∣ E)$ ; the posterior probabilities over an exhaustive set must add to a fixed number.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Let $E_{1}, E_{2}, E_{3}$ be the events that Jayant, Sonia, Oliver processed the form, and $E$ the event of an error. $P (E_{1}) = \frac{5}{10}, P (E_{2}) = \frac{1}{5}, P (E_{3}) = \frac{3}{10}$ ; $P (E ∣ E_{1}) = 0.06, P (E ∣ E_{2}) = 0.04, P (E ∣ E_{3}) = 0.03$ .
(i) $P (E \cap E_{2}) = P (E_{2}) P (E ∣ E_{2}) = \frac{1}{5} \times 0.04 = 0.008$ .
1 mark
(ii) By the theorem of total probability: $P (E) = 0.5 \times 0.06 + 0.2 \times 0.04 + 0.3 \times 0.03 = 0.03 + 0.008 + 0.009 = 0.047$ .
1 mark
(iii)(a) By Bayes' theorem: $P (E_{1} ∣ E) = \frac{P ( E _{1} ) P ( E ∣ E _{1} )}{P ( E )} = \frac{0.03}{0.047} = \frac{30}{47}$ .
1 mark
(iii)(a) Required probability $P (\overset{ˉ}{E}_{1} ∣ E) = 1 - \frac{30}{47} = \frac{17}{47}$ .
1 mark
(iii)(b) $\sum_{i = 1}^{3} P (E_{i} ∣ E) = \frac{P ( E \cap E _{1} ) + P ( E \cap E _{2} ) + P ( E \cap E _{3} )}{P ( E )} = \frac{P ( E \cap ( E _{1} \cup E _{2} \cup E _{3} ))}{P ( E )}$ , since $E_{1}, E_{2}, E_{3}$ are mutually exclusive.
1 mark
(iii)(b) As $E_{1} \cup E_{2} \cup E_{3} = S$ (exhaustive), this equals $\frac{P ( E )}{P ( E )} = 1$ . (Check: $\frac{30}{47} + \frac{8}{47} + \frac{9}{47} = 1$ .)
1 mark

Answer: (i) 0.008; (ii) 0.047; (iii)(a) 17/47; (iii)(b) 1

Where the marks are

(i) $P (E_{2}) P (E ∣ E_{2}) = 0.008$ (1 mark)
(ii) Total probability $P (E) = 0.047$ (1 mark)
(iii)(a) Bayes' theorem gives $P (E_{1} ∣ E) = 30/47$ (1 mark)
(iii)(a) Not Jayant: $1 - 30/47 = 17/47$ (1 mark)
(iii)(b) Uses mutually exclusive and exhaustive events to combine the sum (1 mark)
(iii)(b) Sum of posterior probabilities $= 1$ (1 mark)

Revise the chapter

Chapter: Probability (notes, formulas, where marks are lost and every past question)
Topic: Conditional Probability

Practise this chapterBoard-style questions from Probability, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Mathematics sample paper 2023-24: question 36

Hints

Step-by-step solution

Revise the chapter

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