Listing Sample SpacesDate: 02/13/2003 at 11:45:29 From: Diann Subject: Equally Likely Outcomes An urn contains 2 red, 2 black and 1 green ball, from which a sample of two is chosen without replacement. Determine a sample space of equally likely outcomes and assign probabilities. My problem with this is that when you assign the probabilities that the sample space for red is (1,2 1,3 1,4 1,5 2,3 2,4 2,5), the sample space for blue is (3,4 3,5 4,5), but because 2 balls are being drawn there can be no sample space for green; you can't draw 2 green balls because there is only one. What does this mean, that there is 0 probability for green? Date: 02/18/2003 at 22:20:20 From: Doctor Wallace Subject: Re: Equally Likely Outcomes Hello Diann, Remember that a "sample space" is just another name for "all possible outcomes for an event." In your example, the urn contains 5 balls - 2 red, 2 black, 1 green. Now, before we can have a sample space, we have to have an event. For example, if I said the event was "Selecting 1 ball from the urn," then the sample space would contain the following: red, red, black, black, green So the probability of choosing 1 red ball would be 2/5, since 2 outcomes are favorable, and the sample space contains 5 possibilities. Your example said you are to draw 2 balls, without replacement. That is enough to determine the sample space of outcomes for this event. There is no separate "sample space for red" or "sample space for green." There is only ONE sample space for "drawing 2 balls without replacement" - and it contains a list of ALL possible outcomes for this event. To make it easier to list them all, think of the red balls and the black balls with numbers on them so we can distinguish them. Red1, Red2, Black1, Black2. Here it is: First pick Second pick ----------------- --------------- (1) red1 red2 (2) red2 red1 (3) red1 black1 (4) red1 black2 (5) red2 black1 (6) red2 black2 (7) red1 green (8) red2 green (9) black1 red1 (10) black1 red2 (11) black2 red1 (12) black2 red2 (13) black1 black2 (14) black2 black1 (15) black1 green (16) black2 green (17) green red1 (18) green red2 (19) green black1 (20) green black2 The advantage of a list like this is that each of these 20 outcomes is equally likely to occur. This is why it is VITAL that we distinguish between the 2 red balls and the 2 green ones. If we don't, we miss some of the possibilities. Now, armed with the chart, you can go through and make a list of the probabilities of various things. I think this is what the "assign probabilities" part of your question means. For example, what is the probability that you draw green, then black in that order? Well, we know there are 20 possible outcomes, so the denominator of the probability fraction is 20. The numerator will be the number of "favorable outcomes," or how many ways we can draw green, then black. Looking at the chart, green then black is represented by numbers (19) and (20). This is 2 chances out of 20, or 2/20, which we will reduce to 1/10. So the probability of drawing green then black is 1/10. You can use the sample space chart to answer ANY question you want regarding this problem now. Let's do one more. What is the probability that you draw red then black? In the chart, this is represented by numbers (3), (4), (5), and (6). This is 4 chances out of 20, or 4/20 which is 1/5. Finally, to answer the question you mentioned about green, the probability that you draw green then green is indeed zero. This is because there is only 1 green in the urn, so there are NO entries in the sample space that contain green then green. So there are 0 chances out of 20 or 0/20 which, of course, is 0. The strategy of listing the sample space will work for probability problems where the space is small and manageable, like this one. Other problems have sample spaces with thousands or even millions of possibilities, and to solve those we must turn to the formulas. I hope this helps you, Diann. Thanks for writing to Dr. Math! - Doctor Wallace, The Math Forum http://mathforum.org/dr.math/ |
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