I have but one question to ask of you.
How far can a dog run into the woods?
Here are some answers to the last quiz:
1. What is the probability that no lightbulbs will go out?
The probability that any one lightbulb will not go out is .95 (which is 1.00 - .05). The probability that ten out of ten lightbulbs will survive the night would be (.95) (.95) (.95) (.95) (.95) (.95) (.95) (.95) (.95) (.95) = .60 or 60%
2. What is the probability that any one lightbulb will go out?
The probability that any one of the lightbulbs (it doesn’t matter which) will go out in the group of ten would be found by dividing .05 [the probability that any one single lightbulb goes out] by .60 [the probability that none of the lightbulbs go out], which would equal .08 or 8%. The mistake that many people make in working with this problem is that they will multiply (.95) times itself nine times, and then multiply that result by .05. This does not work, because each lightbulb operates independently of the others. The answer that you would obtain that way would be smaller than .05, which clearly would not make sense – intuitively, you realize that the answer doesn’t make sense if it is less than .05. Intuition is an important tool!
3. What is the probability that any two lightbulbs will go out (together)?
The probability that any two lightbulbs would go out together, i.e., jointly, would be (.05) (.05) or .0025, or .25%. Using the same reasoning as in the solution to the second problem above, we would obtain our answer by dividing .0025 by .60. The solution would be .000416 or .0416%. Intuitively, this also makes sense because the probability of two lightbulbs going out together should be smaller than the probability of only one lightbulb going out in the whole group.
4. What is the probability that up to three lightbulbs will go out?
This question is asking for the sum of the probabilities that: one lightbulb will go out alone; that two lightbulbs will go out together; and that three lightbulbs will go out together. To solve this one, we add the solution to the second problem (.08), to the solution to the third problem (.025), and then add (.05) (.05) (.05)/ (.60) [the probability of three bulbs going out together], and we get: .08 + .000416 + .000208 = .080624 or 8.06%. Intuitively this makes sense because the additive probabilities of all three cases would have to be a number greater than any one of the individual numbers.
5. What is the probability that either three or four lightbulbs will go out?
I am feeling exhausted and even a bit cantankerous. I’ll let you figure this one out yourself, while I have coffee. :)