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Q: Probability of selecting balls from an urn. ( Answered ,   1 Comment )
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 Subject: Probability of selecting balls from an urn. Category: Science > Math Asked by: wobbleone-ga List Price: \$10.00 Posted: 10 Sep 2006 17:25 PDT Expires: 10 Oct 2006 17:25 PDT Question ID: 763991
 ```An urn contains one red ball and four with balls. You and a friend take turns selecting a ball from the urn, one at a time. The first person to select the red ball wins. If you draw first, what is the probability that you will win if sampling is done with replacement and the probability without replacement?```
 ```Hello! Let's start with the case in which the sampling is done without replacement. Since there is a total of 5 balls, you can draw balls a maximum of three times: in the first turn, in the third turn (if your friend doesn't win in the 2nd one) and in the fifth turn (if your friend doesn't win in the 2nd or 4th one). Therefore, we can write the probability of winning as: P(Win) = P(Win in 1st turn) + P(Win in 3rd turn) + P(Win in 5th turn) In the first turn, there's one red ball out of a total of 5 balls. Therefore, the probability of winning in this turn is 1/5 = 0.2: P(Win in 1st turn) = 0.2 Now, in order to win in the 3rd turn, it must happen that: - You draw a white ball in the 1st turn - Your friend draws a white ball in the 2nd turn - You draw the red ball in the third turn The probability that you draw a white ball in the 1st turn is clearly 4/5 = 0.8. After this turn, there are 4 balls left, one of which is red. Therefore, the probability that your friend draws a white ball in the 2nd turn is 3/4 = 0.75. In the third turn, there are 3 balls left, one of which is red. Therefore, the probability of drawing the red ball is 1/3 = 0.333... We thus conclude that: P(Win in 3rd turn) = (4/5)*(3/4)*(1/3) = 0.2 Finally, we have to calculate the probability that you win in the 5th turn. Using the same reasoning as above, it must happen that neither you or your friend draw the red ball in the first four turns. Therefore, the probability of this is: P(Win in 5th turn) = (4/5)*(3/4)*(2/3)*(1/2) = 0.2 We conclude that: P(Win) = 0.2 + 0.2 + 0.2 = 0.6 If you draw first, you have a 0.6 probability of winning. Let's now see the case with replacement. In this case, theoretically, there can be any number of turns before someone draws the red ball. Since there is replacement, the probability of drawing the red ball is always 1/5 (as there are always 5 balls), so the probability of drawing a white one is 4/5. The probability of winning can be written as: P(Win) = P(Win in 1st turn) + P(Win in 3rd turn) + P(Win in 5th turn) + + P(Win in 7th turn) + ... Now, the probability of winning in the 1st turn is clearly 1/5. In order to win in the 3rd turn, both you and your friend must draw white balls in the 1st and 2nd turn, and you must draw the red one in the 3rd turn. Therefore, the probability of this event is (4/5)*(4/5)*(1/5) = [(4/5)^2]*(1/5). In order to win in the 5th turn, both you and your friend must draw white balls in the 1st, 2nd, 3rd and 4th turn, and you must draw the red one in the fifth turn. Therefore, the probability of this event is: (4/5)*(4/5)*(4/5)*(4/5)*(1/5) = [[(4/5)^2]^2]*(1/5) The pattern should now be clear. The probability of winning in the 7th turn would be [[(4/5)^2]^3]*(1/5); in the 9th turn it would be [[(4/5)^2]^4]*(1/5), and so on. Therefore, P(Win) = 1/5 + [(4/5)^2]*(1/5) + [[(4/5)^2]^2]*(1/5) + [[(4/5)^2]^3]*(1/5) + + ... P(Win) = (1/5)*(1 + [(4/5)^2] + [[(4/5)^2]^2] + [[(4/5)^2]^3] + ...) The infinite sum in the above expression has the form 1 + r + r^2 + r^3 + ... This sum (as long as |r|<1) is known to converge to 1/(1-r). Therefore, we get: P(Win) = (1/5)/(1 - (4/5)^2) = 5/9 = 0.555... Therefore, the probability of winning if you draw first and the sampling is with replacement is 5/9, a bit smaller than in the other case. I hope this helps! If you have any doubt regarding my answer, please don't hesitate to request clarification before rating it. Otherwise, I await your rating and final comments. Best wishes! elmarto```
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