I cannot get past this question:
Assume that 12 people, including the husband and wife pair, apply for 5 sales positions. People are hired at random. What is the probability that one is hired and one is not?
The sample space is C(12, 5). I tried finding first the probability that one is hired...
Since there are 336 outcomes, which I got by multiplying 8 x 7 x 6, that means that each outcome is assigned a probability of 1/336. Since they were asking for balls with odd numbers, I "removed" the even numbers from the original set of outcomes, so that it would be 4 x 3 x 2. So I assumed that...
The final question in my homework says:
Assume the balls in the box are numbered 1 through 8, and that an experiment consists of randomly selecting 3 balls one after another without replacement. What probability should be assigned to the event that at least one ball has an odd number?
I have...
But why is the answer the union of the second and third subsets if the question is asking for either one of them? Shouldn't it be 60? Or am I misinterpreting the question?
I am practicing for my math exam next week and I came across this problem:
A set has 200 elements in it. It is partitioned into three subsets so that the second and third subsets have the same number of elements. If four times the number of elements in the second subset is three times as many...