Question

A defective car has a probability of 2/3 upon turning on the ignition in each attempt. Assume attempts

are independent.

• (a) What is the probability that exactly 3 attempts are needed until the car starts?

• (b) What is the probability that 3 or 4 attempts are needed?

• (c) What is the probability of success in 4 or more trials?

Answer 1

By using multiplication theorem of probability of independent event, we will solve.

Probability that car starts in 1 attempt= probability of turning on to ignition temperature=P=2/3

So, probability that car not starts in a attempt=q=(1-p)=1-2/3=1/3

a) Ans- the car starts in exactly three attempts if in the first two attempts it doesn't starts. So, by using multiplication theorem of probability of independent event.

This, probability that exactly three attempts are needed until the car starts=(1/3)*(1/3)*(2/3)=2/27

b) Ans- probability that 3 or 4 attempts needed=probability that 3 attempts needed+probability that 4 attempts needed=(1/3*1/3*2/3)+(1/3*1/3*1/3*2/3)=2/27+2/81=8/81

c) Ans-probability of success in 4 or more trials=1- probability of success in less than or equal to 3 trials=1- {p(success in 1 trial)+P(success in 2 trials)+P(probability of success in 3 trials)}=

1-{(2/3)+(1/3*2/3)+(1/3*1/3*2/3)}=1-(2/3+2/9+2/27)=1-(26/27)=1/27