A box contains four $40$ W, five $60$ W and six $75$ W bulbs. If the bulbs are chosen one by one in random order, what is the probability that at least two bulbs should to be selected to obtain one of $75$ W?
A. $9/15$
B. $6/15$
C. $3/15$
D. $1/15$
I have thought about the following but I do not know if it is correct:
The probability that one of $75$W is selected in the second selection is $\frac{9}{15}\frac{6}{14}$
The probability that one of $75$W is selected in the third election is $\frac{9}{15}\frac{8}{14}\frac{6}{13}$.
and so on, I think that the probability that they are asking me is the sum of all the previous probabilities because they ask me the probability that at least two balls are selected to select one of P. I am reasoning well? Is there an easier way to solve the problem? Thank you very much
Observe that the probability that it takes at least two tries to obtain a 75 W bulb means that a 75 W bulb was not selected on the first try. Since six of the $4 + 5 + 6 = 15$ bulbs are 75 Watt bulbs, the probability of selecting a 75 W bulb on the first try is $$\Pr(\text{75 W bulb selected on first try}) = \frac{6}{15}$$ Hence, the probability that it takes at least two tries to obtain a 75 W bulb is $$\Pr(\text{at least two attempts to obtain 75 W bulb}) = 1 - \frac{6}{15} = \frac{9}{15}$$ Your idea of adding the probabilities that the first 75 W bulb is selected on the second, third, fourth, fifth, sixth, seventh, eighth, ninth, or tenth tries is sound. However, it involves many more calculations and, with them, many more opportunities to make an error.