In a lot of $50$ lightbulbs, there are $2$ bad bulbs. How many bulbs should be examined so that the probability of finding at least $1$ bad bulb is at least $1/2$?
My effort:
Suppose $n$, where $0 \leq n \leq 50$, bulbs are examined. Then the probability of finding no bad bulb is $$ \frac{ 48 \choose n }{ 50 \choose n}.$$ So the probability of finding at least $1$ bad bulb is $$ 1 – \frac{ 48 \choose n }{ 50 \choose n}.$$ Thus we must have $$ 1 – \frac{ 48 \choose n }{ 50 \choose n} \geq \frac{1}{2}.$$
Am I proceeding in a correct fashion? And if so, what would be our desired value of $n$?
Best Answer
Your formula is correct. I suggest you now try to simplify it using $${r\choose k} = \frac{r!}{(r-k)!k!}$$