A technician working at quality control department of a production plant is inspecting a lot of $1000$ light bulbs, among which are $20$ defective. He chooses two light bulbs randomly from the lot without replacement.
$$X_1 = \begin{cases} 1, & \text{if the first light bulb is defective;} \\ 0, & \text{otherwise.}\end{cases}$$
$$X_2= \begin{cases} 1, & \text{if the second light bulb is defective;} \\ 0, & \text{otherwise.}\end{cases}$$
Find the probability that at least one light bulb chosen is defective.
I did this question like
$$P(X_1 = 1, X_2 = 0) + P(X_1 = 0, X_2 = 1) + P(X_1 = 1, X_2 = 1) \\ = (20/1000 \cdot 980/999) + (980/1000 \cdot 20/999) + (20/1000 \cdot 19/999)$$
but I am not sure that it's true.
Best Answer
The calculation looks correct.
Alternately, the probability we are looking for is $1$ minus the two bulbs are both good. Using your procedure, we find that the required probability is $$1-\left(\frac{980}{1000}\right)\left(\frac{979}{999}\right).$$
There are alternative ways to calculate the probability that both bulbs are good. We can choose $2$ bulbs from the $1000$ in $\dbinom{1000}{2}$ equally likely ways. The number of ways to choose $2$ good ones is $\dfrac{980}{2}$. Divide.
There are also various ways to approximate the required probability. If you have met the Poisson approximation, note that the number of bad bulbs in a sample of $2$ has approximately Poisson distribution, with $\lambda=2\frac{20}{1000}$. So the probability of no bad is roughly $e^{-\lambda}$.