(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Step-by-Step Solution
Key Concept: Use the definition of probability as \(P(E)=\dfrac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}\). For part (ii) apply the concept of conditional probability when sampling without replacement.
### Part (i)
1. Total number of bulbs in the lot = 20.
2. Number of defective bulbs = 4.
3. Since each bulb is equally likely to be drawn, the probability that the drawn bulb is defective is
$$\displaystyle P(\text{defective}) = \frac{\text{defective bulbs}}{\text{total bulbs}} = \frac{4}{20} = \frac{1}{5}=0.2.$$
### Part (ii)
1. The first bulb drawn is not defective. Hence one non‑defective bulb is removed from the lot.
2. Remaining bulbs = 20 – 1 = 19.
3. Original non‑defective bulbs = 20 – 4 = 16. After removing one non‑defective bulb, the remaining non‑defective bulbs = 16 – 1 = 15.
4. The number of defective bulbs is unchanged (still 4).
5. Probability that the second drawn bulb is not defective (given the first was not defective and not replaced) is
$$\displaystyle P(\text{not defective}\mid \text{first not defective}) = \frac{\text{remaining non‑defective bulbs}}{\text{remaining total bulbs}} = \frac{15}{19}.$$
6. If a decimal answer is required, \(\frac{15}{19} \approx 0.7895\) (rounded to four decimal places).
Correct Answer: (i) \(\dfrac{1}{5}\) (or 0.2)\n(ii) \(\dfrac{15}{19}\) (approximately 0.7895)