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.

1. Find the probability that Sonia processed the form and committed an error.
2. Find the total probability of committing an error in processing the form.
3. 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.
   OR
   Let E be the event of committing an error in processing the form and let E₁, E₂ and E₃ be the events that Jayant, Sonia and Oliver processed the form. Find the value of `sum_(i = 1)^3P(E_i|E)`.

Answer 1

Let E₁, E₂, E₃ be the events that Jayant, Sonia and Oliver processed the form, which are clearly pairwise mutually exclusive and exhaustive set of events.

Then P(E₁) = `50/100 = 5/10`, P(E₂) = `20/100 = 1/5` and P(E₃) = `30/100 = 3/10`.

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

We have, P(E | E₁) = 0.06, P(E | E₂) = 0.04, P(E | E₃) = 0.03.

i. The probability that Sonia processed the form and committed an error is given by

P(E ∩ E₂) = P(E₂).P(E | E₂)

= `1/5 xx 0.04`

= 0.008.

ii. The total probability of committing an error in processing the form is given by

P(E) = P(E₁).P(E | E₁) + P(E₂).P(E | E₂) + P(E₃).P(E | E₃)

P(E) = `50/100 xx 0.06 + 20/100 xx 0.04 + 30/100 xx 0.03`

= 0.047.

iii. The probability that the form is processed by Jayant given that form has an error is given by

P(E₁ | E) = `(P(E|E_1) xx P(E_1))/(P(E|E_1).P(E_1) + P(E|E_2).P(E|E_3).P(E_3)`

= `(0.06 xx 50/100)/(0.06 xx 50/100 + 0.04 xx 20/100 + 0.03 xx 30/100)`

= `30/47`

Therefore, the required probability that the form is not processed by Jayant given that form has an error = `P(overlineE_1|E) = 1 - P(E_1 | E) = 1 - 30/47 = 17/47`.

OR

`sum_(i = 1)^3 P(E_i | E) = P(E_1 | E) + P(E_2 | E) + P(E_3 | E)` = 1

Since, sum of the posterior probabilities is 1.

(We have,  `sum_(i = 1)^3 P(E_i | E) = P(E_1 | E) + P(E_2 | E) + P(E_3 | E)`

= `(P(E ∩ E_1) + P(E ∩ E_2) + P(E ∩ E_3))/(P(E))`

= `(P((E ∩ E_1) ∪ (E ∩ E_2) ∪ (E ∩ E_3)))/(P(E))` as E_i and E_j ; i ≠ j, are mutually exclusive events

= `(P(E ∩ (E_1 ∪ E_2 ∩ E_3)))/(P(E)) = (P(E ∩ S))/(P(E)) = (P(E))/(P(E))` = 1; 'S' being the sample space)