A box in a certain supply room contains three 40-w lightbulbs, five 60-w bulbs, and seven 75-w bulbs. suppose that three bulbs are randomly selected. (round your answers to four decimal places.) (d) suppose now that bulbs are to be selected one by one until a 75-w bulb is found. what is the probability that it is necessary to examine at least six bulbs?

If we are selecting bulbs one at a time, and we're looking for the probability that we don't find a bulb until at least the sixth try, we didn't find the right bulb on the first 5 tries. On the 6th try, we might have found the right bulb, or maybe we didn't and needed more tries, but we know that we must have missed on the first 5 tries.

So, we want the following probability:
[tex]P({1^{st}}_{\text{wrong}})*P({2^{nd}}_{\text{wrong}})*P({3^{rd}}_{\text{wrong}})*P({4^{th}}_{\text{wrong}})*P({5^{th}}_{\text{wrong}})*P({6^{th}}_{\text{right or wrong}})[/tex]

At any point, the probability of choosing the wrong bulb is [tex]\frac{\text{\# wrong bulbs left}}{\text{\# bulbs left}}[/tex]

Also, at any point, the probability of choosing a right or a wrong bulb is [tex]\frac{\text{\# bulbs left}}{\text{\# bulbs left}}[/tex] which is equivalent to 1 . If you choose a bulb, you either chose a right one, or a wrong one, there's no other choice.

So, the probability we're looking for is as follows:

[tex]P({1^{st}}_{\text{wrong}})*P({2^{nd}}_{\text{wrong}})*P({3^{rd}}_{\text{wrong}})*P({4^{th}}_{\text{wrong}})*P({5^{th}}_{\text{wrong}})*P({6^{th}}_{\text{right or wrong}})[/tex][tex]\frac{8}{15}*\frac{7}{14}*\frac{6}{13}*\frac{5}{12}*\frac{4}{11}*\frac{10}{10}[/tex]

0.0186480186

So, there is a 1.8648% chance of picking the correct bulb on the 6th try of later. In terms of probability, the probability (to 4 decimal places) is 0.0186