Posted by Kromey at 3:11am Sep 19 '07Here's a statistics scenario that boggles the minds of even the most seasoned student of mathematics, although we need some build-up first. I put it on the Debates board because, well, knowing y'all like I do (and seeing the lively statistics debate on Community) this one will spawn a good debate.
5 coin flips:
Assuming a fair coin (i.e. probability of heads is exactly 50%), what is the probability of 5 heads out of 5 flips?
Solution:
Since each coin flip is an independent trial, each flip of the coin has a 1/2 chance of being heads; therefore, to get 5 heads, the probability is (1/2)^5 = 1/32 = 0.03125
Choosing 5 marbles from a bag of 10:
Let's say you have a bag of 10 marbles; 5 red and 5 black. You want to choose 5 marbles from this bag. Each time you pick a marble, you record which color you picked and then put it back. What is the probability of picking 5 red?
Solution:
Since we are choosing with replacement (the statistician's way of saying "we pick one then put it back"), each marble is an independent trial. Therefore, the probability of getting a red marble is 5/10 = 1/2; this makes the probability of getting 5 red marbles the same as that of getting 5 heads on a fair coin, 1/32.
5 marbles from a bag of 10, again:
Same as the previous, only this time you don't put the marble back after you pull one out. What's the probability of getting 5 red ones now?
Solution:
This one is a bit tricky, because each time you pick a marble the probability for the next pick changes. For example, if your first marble is red, then the probability of the second being red is 4/9 (because there are 4 red marbles left out of 9 total marbles left). So the solution becomes 1/2 * 4/9 * 3/8 * 2/7 * 1/6 = 24/6048 = 1/252.
Many students throw up their hands when the professor presents this solution and ask, "But what about all those times that you pick a black one?" Well, you have a 251/252 chance of picking at least one black one; you could enumerate all possible choices of 5 marbles, but you will end up with this result (assuming you don't make a mistake). Don't believe me? Try it! (You may want to try it with a smaller sample, maybe only 4 marbles.)
As confusing as the previous one is for many statistics students, the next one is the real killer:
Choose 5 marbles from a bag of 10, without replacement, and without looking:
Same scenario as above, that is you have a bag of 5 red and 5 black marbles. You want to pick 5 marbles, and you are not going to put them back after you pick them. The catch is that you will not look at the marble as you pull it out. What is the probability that you will pull out 5 red marbles?
Solution:
Tune in tomorrow for the exciting solution to this riddle!
Hint: The famous Schrödinger's cat thought experiment is based on the same principle.
5 coin flips:
Assuming a fair coin (i.e. probability of heads is exactly 50%), what is the probability of 5 heads out of 5 flips?
Solution:
Since each coin flip is an independent trial, each flip of the coin has a 1/2 chance of being heads; therefore, to get 5 heads, the probability is (1/2)^5 = 1/32 = 0.03125
Choosing 5 marbles from a bag of 10:
Let's say you have a bag of 10 marbles; 5 red and 5 black. You want to choose 5 marbles from this bag. Each time you pick a marble, you record which color you picked and then put it back. What is the probability of picking 5 red?
Solution:
Since we are choosing with replacement (the statistician's way of saying "we pick one then put it back"), each marble is an independent trial. Therefore, the probability of getting a red marble is 5/10 = 1/2; this makes the probability of getting 5 red marbles the same as that of getting 5 heads on a fair coin, 1/32.
5 marbles from a bag of 10, again:
Same as the previous, only this time you don't put the marble back after you pull one out. What's the probability of getting 5 red ones now?
Solution:
This one is a bit tricky, because each time you pick a marble the probability for the next pick changes. For example, if your first marble is red, then the probability of the second being red is 4/9 (because there are 4 red marbles left out of 9 total marbles left). So the solution becomes 1/2 * 4/9 * 3/8 * 2/7 * 1/6 = 24/6048 = 1/252.
Many students throw up their hands when the professor presents this solution and ask, "But what about all those times that you pick a black one?" Well, you have a 251/252 chance of picking at least one black one; you could enumerate all possible choices of 5 marbles, but you will end up with this result (assuming you don't make a mistake). Don't believe me? Try it! (You may want to try it with a smaller sample, maybe only 4 marbles.)
As confusing as the previous one is for many statistics students, the next one is the real killer:
Choose 5 marbles from a bag of 10, without replacement, and without looking:
Same scenario as above, that is you have a bag of 5 red and 5 black marbles. You want to pick 5 marbles, and you are not going to put them back after you pick them. The catch is that you will not look at the marble as you pull it out. What is the probability that you will pull out 5 red marbles?
Solution:
Tune in tomorrow for the exciting solution to this riddle!
Hint: The famous Schrödinger's cat thought experiment is based on the same principle.
Coins and marbles - Kromey