Question

Three persons viz. Amber, Bonzi and Comet are manufacturing cars which run on petrol and on battery as well. Their production share in the market is 60%, 30% and 10% respectively. Of their respective production capacities, 20%, 10% and 5% cars respectively are electric (or battery operated).

Based on the above, answer the following questions:

1. 1. What is the probability that a randomly selected car is an electric car?   [2]
   2. What is the probability that a randomly selected car is a petrol car?   [2]
2. A car is selected at random and is found to be electric. What is the probability that it was manufactured by Comet?   [1]
3. A car is selected at random and is found to be electric. What is the probability that it was manufactured by Amber or Bonzi?   [1]

Answer 1

(i) (a) Let A → manufacturing by Amber

B → manufacturing by Bonzi

C → manufacturing by Comet

E → manufacturing by Electric car

`P(A) = 60/100`

`P(B) = 30/100`

`P(C) = 10/100`

`P(E//A) = 20/100`

`P(E//B) = 10/100`

`P(E//C) = 5/100`

P(Electric car) = P(A) P(E/A) + P(B) P(E/B) + P(C) P(E/C)

= `60/100 xx 20/100 + 30/100 xx 10/100 + 10/100 xx 5/100`

= `1200/10000 + 300/10000 + 50/10000`

= `(1200 + 300 + 50)/10000`

= `1550/10000`

= 0.155 or 15.5%

OR

(i) (b) P(petrol car) = 1 – P(electric car)

=1 – 0.155 

= 0.845 or 84.5%

(ii) Using Bayes theorem

P(C/E) = `(P(C) P(E//C))/(P(A) P(E//A) + P(B) P(E//B) + P(C) P(E//C)`

= `(10/100 xx 5/100)/(1550/10000)`

= `50/1550`

= `1/31`

(iii) P(C/E) = `(P(A) P(E//A))/(P(A) P(E//A) + P(B) P(E//B) + P(C) P(E//C)`

= `(60/100 xx 20/100)/(1550/10000)`

= `1200/1500`

= `24/31`

P(C/E) = `(P(A) P(E//A))/(P(A) P(E//A) + P(B) P(E//B) + P(C) P(E//C)`

= `(30/100 xx 10/100)/(1550/10000)`

= `300/1550`

= `6/31`

P(Amber or Bonzi) = `24/31 + 6/31`

= `30/31`