Posted: 10/2/2007 10:58:49 AM EDT
| If 5 people (A B C D E) are in an election and two will be selected for a position what are the odds that B or D will be selected? Can anyone do this? Would to answer be 2 to 3? |
|
Select TWO out of FIVE in one choice: A B C D E x x 0 0 0 x 0 x 0 0 x 0 0 x 0 x 0 0 0 x 0 x x 0 0 0 x 0 x 0 <--BOTH selected = 1 in 10 chance, equivalent to 10% odds. 0 x 0 0 x 0 0 x x 0 0 0 x 0 x 0 0 0 x x *-*-*-*-*-*-*-*-*-*-*-*-*-*-*-* Pick one of five, then pick another (one in four) One in five = 20% chance Then... one in four = 25% chance Combined chance that either B or D will be chosen is 20% + 25% = 45% |
|
Here's the forumla I' suppose to use to calculate odds. I wouldn't be wondering except when I handed in the test someone else had a different answer and I want to know who is wrong. Pr(E) / 1 - Pr(E) so i had probability of b or d being chosen is .4 (.2 +.2) .4/(1-.4) = .6666666667 or 2/3 which i believe would be 2 to 3 odds. |
 That formula doesn't describe the situation you're talking about. That's the formula for comparative probability, i.e. the probability of the occurance vs. the probability of it not occurring. The general odds are 1/10. |
I am basing it on that the Probability of E which is the event of B or D being chose. |
I think that is wrong. Combined odds are multiplied. .20 x .25 = .04 4% chance, but since the order doesn't matter, there is an 8% chance I think. |
Oops... That is selecting with replacement, which allows for the possibility of the same person being selected for both positions. |
Add the odds to get either one being selected (OR condition). |
Well, that formula and a 40% probability gives a 2/3 answer. I do NOT think that answer correctly solves the situation you described. What are you not telling us? |
I can't remember what else was mentioned in the question that I had.
|
Please Read Actually, you're both wrong if they are selected one at a time. First, if we want both B and D: If we select individually, we are performing an ordered selection. In that case, there are two outcomes that satisfy the event E - B then C, and C then B. The total number of possible outcomes, selecting in order without replacement, its n!/(n-r)!, which in this case is 5!/2! = 60. To get the actual probability, we divide the size of the set of satisfying events by the size of the sample space. |A|/|S| = 2/60 = 1/30. If we just want one of them, there are 14 possible outcomes that satisfy our event E, so |A| = 14. |A| / |S| = 14/30 = 7/15. ------------- If we select both at the same time, an unordered selection, there are (5 choose 2) = 10 possible outcomes. If we just want one, there are 7 outcomes in A, and 7/10 - 70%. If we want both, there is only one outcome in A, that of BD. 1/10 = 10%. ---------------------------- So to sum it up: If both are selected at the same time: B and D: 1/10 B or D: 7/10 If they are selected one at a time: B and D: 1/30 B or D: 7/15 OP, I think some additional wording was in the problem that you aren't remembering here. |
|
Probability.. haven't done that in a few semesters If you pick 1 candidate at a time: 1/5 + 1/5 = 2/5 = .4 Prob of pick B or D 1/4 + 1/4 = 2/4 = .5 .5 x .4 = .20 = 1 to 5 "odds" I think. YMMV I got a D in my "Business math" class (which covered probability partially)  |
|
probability that both are selected. 1/10 basically 1case/5C2 probability that at least one of them are selected = 1 - prob. of neither of them. prob. of neither of them is 6/10 since there are 6 possible combinations of them not getting selected. so at least one of them selected would be 4/10 |
There are only 3 unordered cases in which they aren't selected at all: AC, AE, and CE. Please see my above post for 4 interpretations: both and one for both unordered and ordered selection. |
Nope, can't do it because unlike random selection, an election is (presumably) not random so unless you know something about the electorate's tendencies, you can't determine the mathematical probability of any candidate winning a position. A simple example - the position is US Senator and the candidates are: Hillary, Fred Thompson, and three homeless guys who mutter to themselves. What are the odds that a homeless guy will win? |
He is assuming equal probability for the sake of the problem. We could just as easily be talking about balls from a basket. |
The question was "can anyone do this." The answer is "no" and any mathematics professor would agree. Maybe the point of the problem is to identify those examples in which probabilities can be determined (like throwing fair dice) and those in which the probabilities can't be determined (like elections). |
I took the "can anyone do this" to mean "can someone please do this for me", rather than as part of the problem (much like his followup question, "Would the answer be 2 to 3?"). Also, he states a couple posts down that this is assuming equal probability. |
|
Odds s the ratio of favorable occurrence to UNFAVORABLE occurrence. So if two events out of five are favorable then the odds are 2 to 3 or 2/3 Probability is the ratio of favorable events to the TOTAL number of events and is usualy given as a decimal, not a fraction, so as to avoid confusion with odds. In the above scenario, the probability would be 2 divided by 5 which is 0.40 or 40% CHANCE (assuming equal likelihood) If you want to do 2 events at the same time, from a combination, where order is not important, then you are doing either a combination or a permutation and I forget which. Use a previous post to determine the sample space (all possible outcomes listed) and do odds using the definition I gave in the first paragraph above. |