Question

1. Five defective integrated circuits have accidentally passed the quality control inspection and shipped to a retailer in a batch of a total of 25 integrated circuits. Before selling them, the retailer decides to test the integrated circuits by randomly picking 10 integrated circuits out of the 25 that were shipped.

(a) [10 points] What is the probability that there are exactly two defective integrated circuits in this sample of 10?

(b) [10 points] What is the probability that there are at most two defective integrated circuits in this sample of 10?

Answer 1

We have a total of 10 defective integrated circuits. So a total of 15 integrated circuits are good.

Total number of possible ways of selecting 10 integrated circuits of 25 is : ²⁵C₁₀

a) The probability that there is exactly 2 defective integrated circuits in the sample of 10 is:-

(number of ways of selecting 2 defective ic * number of ways of selecting remaining 8 good ic) / total no of outcomes

i.e. (¹⁰C₂ * ¹⁵C₈) / ²⁵C₁₀

= (45*6435) / 3268760

= 0.0886

=8.9%

So the probability of getting exactly 2 defective integrated circuits is 0.0886 or  8.9%

b)The probability of getting at most 2 defective integrated circuits is:-

Probability of getting 0 defective ic + probability of getting 1 defective ic + probability of getting 2 defective ic

i.e. (¹⁵C₁₀ / ²⁵C₁₀) + ((¹⁰C₁ * ¹⁵C₉) / ²⁵C₁₀) + ((¹⁰C₂ * ¹⁵C₈) / ²⁵C₁₀)

= 0.0009 + 0.0153 + 0.0886

=0.1048

=10.5%

Therefore, the probability of getting atmost 2 defective integrated circuits is 0.1048 or 10.5%.