In a box containing 15 bulbs, 5 are defective. If 5 bulbs are selected at random from the box, the probability of the event that
(i) none of them is defective is `12/143`
(ii) only one of them is defective is `50/143`
(iii) atleast one of them is defective is `131/143`

Answer

To solve the problem step by step, we will calculate the probabilities for each case as described in the video transcript.

### Given:
- Total bulbs = 15
- Defective bulbs = 5
- Good bulbs = 10 (since 15 - 5 = 10)
- Bulbs selected = 5

### Total Ways to Select 5 Bulbs:
The total number of ways to select 5 bulbs from 15 is given by the combination formula:
\[ \text{Total ways} = \binom{15}{5} \]

Calculating this:
\[ \binom{15}{5} = \frac{15!}{5!(15-5)!} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003 \]

### (i) Probability that none of them is defective:
To find the probability that none of the selected bulbs are defective, we need to select all 5 from the good bulbs (10 good bulbs).

The number of ways to select 5 good bulbs from 10 is:
\[ \text{Ways to select 0 defective} = \binom{5}{0} \times \binom{10}{5} = 1 \times \binom{10}{5} \]
Calculating \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
Thus, the number of ways to select 5 good bulbs is 252.

Now, the probability is:
\[ P(\text{none defective}) = \frac{\text{Ways to select 0 defective}}{\text{Total ways}} = \frac{252}{3003} = \frac{12}{143} \]

### (ii) Probability that only one of them is defective:
To find the probability that exactly one of the selected bulbs is defective, we need to select 1 defective bulb and 4 good bulbs.

The number of ways to select 1 defective bulb from 5 is:
\[ \text{Ways to select 1 defective} = \binom{5}{1} \times \binom{10}{4} \]
Calculating \(\binom{10}{4}\):
\[ \binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
Thus, the number of ways to select 1 defective and 4 good bulbs is:
\[ \text{Ways} = 5 \times 210 = 1050 \]

Now, the probability is:
\[ P(\text{only one defective}) = \frac{1050}{3003} = \frac{50}{143} \]

### (iii) Probability that at least one of them is defective:
To find the probability that at least one bulb is defective, we can use the complement rule. This means we can subtract the probability of selecting none defective from 1.

\[ P(\text{at least one defective}) = 1 - P(\text{none defective}) = 1 - \frac{12}{143} = \frac{143 - 12}{143} = \frac{131}{143} \]

### Final Answers:
- (i) Probability that none of them is defective: \(\frac{12}{143}\)
- (ii) Probability that only one of them is defective: \(\frac{50}{143}\)
- (iii) Probability that at least one of them is defective: \(\frac{131}{143}\)