Question

Assume that light bulbs bought at a supermarket have a probability 1/3 of being defective.
1）If you buy 10 light bulbs, what is the probability that 3 are defective?

2）If you buy one light bulb per day, what is the probability that the first defective one is on day three?
3）If you buy one light bulb per day, and the one bought on day one was defective, what is the probability that the next defective one is on day 5?

Answer 1

Concept

This is a binomial distribution problem as

1. A random bought bulb is either defective or not defective .

2. The probability of a random bulb of being defective remains constant for each trial which is 1/3

3. Each trial is independent of each other.

4. The number of trials is fixed in each case.

A binomial probability of k successes in n trials is given by P(X = k) = nCk ×(p)k ×(1-p)n-k

Calculation

Let X be the number of defective bulbs bought. with p = 1/3 , q = 1- p = 2/3

A) probability of buying 3 defective bulbs out of 10 bulbs

= P( X = 3)

= 10​​​​​​C3 ×(1/3)3 × (2/3)7

= 0.26

B) here order of success is fixed

So, we will break it into 2 probabillities

Hence, probability that first defective bulb is bought on day 3

= (probability that no defective bulbs is bought in 2 days) × ( probability of buying a defective bulbs on 3rd day)

=(  2​​​​​​C0 × (1/3)0 × (2/3)2 ) × (1/3)

= 4/27

= 0.14815

= 0.148

C)

Similar to case b here also the order of success is defined so we will break the probability here into three different event probabillities.

Hence, probability that first buy is defective and the next defective is on the 5th buy

= (probability that 1st buy is defective ) ×( probabillity that out of next 3 buys no defective bulb is found ) × (probability that 5th buy is defective)

= (1/3) × (3​​​​​​C0 × (1/3)0 × (2/3)3 ) × (1/3)

= (1/3)×(2/3)3 ×(1/3)

= 8/243

= 0.0329

= 0.033