Question 29 - CBSE Class 12 Sample Paper for 2019 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

Question 29

The members of a consulting firm rent cars from three rental agencies: 50% from agency X, 30% from agency Y and 20% from agency Z. From past experience it is known that 9% of the cars from agency X need a service and tuning before renting, 12% of the cars from agency Y need a service and tuning before renting and 10% of the cars from agency Z need a service and tuning before renting. If the rental car delivered to the firm needs service and tuning, find the probability that agency Z is not to be blamed.

Transcript

Question 29 The members of a consulting firm rent cars from three rental agencies: 50% from agency X, 30% from agency Y and 20% from agency Z. From past experience it is known that 9% of the cars from agency X need a service and tuning before renting, 12% of the cars from agency Y need a service and tuning before renting and 10% of the cars from agency Z need a service and tuning before renting. If the rental car delivered to the firm needs service and tuning, find the probability that agency Z is not to be blamed.
Let us define events as
A: Car needs service
E1: Car is rented from agency X
E2: Car is rented from agency Y
E3: Car is rented from agency Z
We need to find the probability that
Car is not chosen from agency Z, if the car needs service
Here, we find probability that
Car is chosen from agency Z, if the car needs service
i.e. P(E3|A)
Here,
P(E1) = Probability that agency X is chosen
= 50% = 50/100
P(E2) = Probability that agency Y is chosen
= 30% = 30/100
P(E3) = Probability that agency Z is chosen
= 20% = 20/100
P(A|E1) = Probability that car needs service if agency X is chosen
= 9% = 9/100
P(A|E2) = Probability that car needs service if agency Y is chosen
= 12% = 12/100
P(A|E3) = Probability that car needs service if agency Z is chosen
= 10% = 10/100
Now,
P(E3|A) = (𝑃(𝐸_3 ).𝑃(𝐴|𝐸_3))/(𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1)+𝑃(𝐸_2 ).𝑃(𝐴|𝐸_2)+𝑃(𝐸_3 ).𝑃(𝐴|𝐸_3))
= (20/100 × 10/100 )/(50/100 × 9/100 + 30/100 × 12/100 + 20/100 × 10/100)
= (20 × 10 )/(50 × 9 + 30 × 12 + 20 × 10)
= 200/(450 + 360 + 200)
= 200/1010
= 20/101
Thus,
P(E3|A) = 20/101
Now,
P(Car is not chosen from agency Z, if the car needs service)
= 1 – P(E3|A)
= 1 – 20/101
= 81/101
Thus, required probability is 𝟖𝟏/𝟏𝟎𝟏

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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