Three light bulbs are selected at random from 20 bulbs of which 5 are defective. What is the probability that (i) none of the bulbs are defective and (ii) exactly one is defective?

Steps to Solve

1. Calculate the total number of ways to choose 3 bulbs out of 20

   Since the order of selection doesn't matter, we use combinations. The total number of ways to choose 3 bulbs out of 20 is given by $ {20 \choose 3} $. $$ {20 \choose 3} = \frac{20!}{3!(20-3)!} = \frac{20!}{3!17!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 10 \times 19 \times 6 = 1140 $$

2. Calculate the number of ways to choose 3 non-defective bulbs

   There are $20 - 5 = 15$ non-defective bulbs. The number of ways to choose 3 non-defective bulbs from these 15 is given by $ {15 \choose 3} $. $$ {15 \choose 3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455 $$

3. Calculate the probability of selecting 3 non-defective bulbs

   The probability of selecting 3 non-defective bulbs is the number of ways to choose 3 non-defective bulbs divided by the total number of ways to choose 3 bulbs. $$ P(\text{none defective}) = \frac{{15 \choose 3}}{{20 \choose 3}} = \frac{455}{1140} = \frac{91}{228} $$

4. Calculate the number of ways to choose exactly 1 defective bulb and 2 non-defective bulbs

   We need to choose 1 defective bulb out of 5 and 2 non-defective bulbs out of 15. The number of ways to do this is given by $ {5 \choose 1} \times {15 \choose 2} $. $$ {5 \choose 1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = 5 $$ $$ {15 \choose 2} = \frac{15!}{2!(15-2)!} = \frac{15!}{2!13!} = \frac{15 \times 14}{2 \times 1} = 15 \times 7 = 105 $$ So, the number of ways to choose 1 defective and 2 non-defective bulbs is $ 5 \times 105 = 525 $.

5. Calculate the probability of selecting exactly 1 defective bulb

   The probability of selecting exactly 1 defective bulb is the number of ways to choose 1 defective bulb and 2 non-defective bulbs divided by the total number of ways to choose 3 bulbs. $$ P(\text{exactly one defective}) = \frac{{5 \choose 1} \times {15 \choose 2}}{{20 \choose 3}} = \frac{525}{1140} = \frac{105}{228} $$