ecn-ec:ecn-pc:ecn-ed:vu@pur-ee.UUCP (09/16/83)
Let p = .5 be the probability that a phenomenum will happen to an item. Thus the probability that it won't happen is 1-p = .5 So with a population of 2, the probability that the phenomenum happens to at least one of the items will be p+p = 1 , while the probability that the phenomenum will not happen to either is (1-p)(1-p) = .25 hence making a total of 1.25 ?!?! Where did I go wrong ? Hao-Nhien Vu (pur-ee!norris)
dixon@ihuxa.UUCP (09/18/83)
For a population of 2 and the probability that something will happen to any of the members being 1/2 (and assuming independent probabilities), then the probability that something will happen to at least one of the members is P(something will happen to member 1 but not member 2) + P(something will happen to member 2 but not member 1) + P(something will happen to both members 1 and 2) = .25 + .25 + .25 = .75 (rather than 1 as claimed). d a dixon ihuxa!dixon
kwmc@hou5d.UUCP (K. W. M. Cochran) (09/19/83)
Let p = .5 be the probability that a phenomenum will happen to
an item. Thus the probability that it won't happen is 1-p = .5
So with a population of 2, the probability that the phenomenum
happens to at least one of the items will be p+p = 1 , while the probability
that the phenomenum will not happen to either is (1-p)(1-p) = .25
hence making a total of 1.25 ?!?!
Where did I go wrong ?
Hao-Nhien Vu (pur-ee!norris)
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Your slip up was when you assumed that the probability that the phenomenum
happens to at least one item is p+p. This probability is 1-(p*p) = .75
Ken Cochran hou5d!kwmcprodeng@fluke.UUCP (Jim Hirning) (09/19/83)
>Let p = .5 be the probability that a phenomenum will happen to >an item. Thus the probability that it won't happen is 1-p = .5 >So with a population of 2, the probability that the phenomenum >happens to at least one of the items will be p+p = 1 , while the probability >that the phenomenum will not happen to either is (1-p)(1-p) = .25 >hence making a total of 1.25 ?!?! >Where did I go wrong ? >Hao-Nhien Vu (pur-ee!norris) The problem lies in your calculation of the first probability to be 1. Obviously this cannot be true (you cannot guarantee to get a "heads" by flipping two coins). The way to calculate this probability: for the phenomenum to happen to "at least one" of the items, it must 1) happen to exactly one of them or 2) happen to both of them. Calling the two items A and B, and the phenomenum happening T, and not happening F, there are four possible outcomes. 1) A T, B T; 2) A T, B F; 3) A F, B T; 4) A F, B F. There are two ways (cases 2 & 3) of the phenomenum happening to exactly one of the items, and one way (case 1) of it happening to both. Since there are four outcomes, the probability of the phenomenum happening to at least one of the items is 2/4 + 1/4 = 3/4. Then the other case, the phenomenum happening to neither, (case 4) has a probability of 1/4. Thus, 3/4 + 1/4 = 1 as required. Debbie Smit