Question

Probability Inquiry!

Answer 1

`A box contains eight 40-W bulbs, nine 60-W bulbs, and seven 75-W bulbs. If bulbs are selected one by one in random order, what is the probability that at least two bulbs must be selected to obtain one that is rated 75 W?`


The correct solution that I obtained was:
\[\text{P}(A)=1-\text{P}(\bar{A})=1-\frac{ 7 }{ 7+9+8 }=0.708\]My question is -- how does the `at least two bulbs` factor into this solution? What if it was at least three or four bulbs?

Answer 2

let P(A) be equal to the probability of getting the bulb on the first try

then 1 - P(A) is equal to the probability of NOT getting the bulb on the first try (AKA, at least 2 bulbs are necessary)

@CShrix does that make sense?

Answer 3

to answer your second question, we would use the same process for 3 or 4 or whatever number of bulbs you want

for example:

the probability of needing at least 3 bulbs = 1 - probability of getting it on the first try - probability of getting it on the second try

Answer 4

Ahh! Yes that makes sense now. Thank you X)