Probabilistic Strategies: Solutions

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1 Probability Victor Xu Probabilistic Strategies: Solutions Western PA ARML Practice April 3, Problems 1. You roll two 6-sided dice. What s the probability of rolling at least one 6? There is a 1 6 probability of rolling a 6 on any individual die. So, there is a 2 6 = 1 3 on rolling a 6 on at least 1 die out of 2. However, we could roll a 6 on both dice! We counted this possibility twice in our 1 3 answer, but we should only count it once. The probability of rolling two sixes is So, the actual probability of rolling at least 1 six is = Two teams are playing each other in a best-to-3 series. Each team has a 50% chance of winning any individual game. What s the probability that the series will go to 5 games? This probability is the same as the probability that in the first four games, both teams won exactly two games. There are ( 4 2) = 6 4-game sequences where this happens, and a total of 2 4 = 16 possible 4-game sequences, so the probability that the series goes to 5 games is 6 16 = You are playing a game against a friend: you and your friend draw the top card off of a deck of cards, and you win if you and your friend both drew cards of the same suit. A deck of 52 cards is shuffled and you start playing, without shuffling the used cards back into the deck. What s the probability that you win the 5 th game? An important observation is that the probability that you win the 5 th game is the same as the probaiblity you win the 1 st game. The probability of winning the first game is since after drawing one card, there are 12 cards left that match the first in suit out of a total of 51 card remaining in the deck. 4. You roll two 6-sided dice. What s the probability that the sum of the numbers you rolled is even? No matter what we rolled on the first roll, the second roll has a 1 2 chance to match the parity of the first roll. The sum if even if and only if the second roll matches the parity of the first roll so the probability the sum is even is 1 2.

2 5. You re playing 3-card anti-poker, where the goal is to not make good poker hands. What s the probability that a random hand of 3 cards from a 52 card deck has cards of not all the same suit and does not contain a pair? We can count the probability that a hand has cards of all the same suit or contains a pair, and take one minus that probability to find the answer. Since it is impossible to have cards of all the same suit and contain a pair, we can add the probability of having cards of all the same suit and the probability of having cards that contain a pair to find the probability that a hand has cards of all the same suit or contains a pair. The probability of having cards of all the same suit can be found by drawing cards one at a time and seeing the probability that the card matches the suit of the previous cards, so we get 12. To find the = probability that our 3 cards contains a pair, we can again find the probability that the 3 cards don t contain a pair and subtract that from 1. This gives us a probability of So, the final probability is 1 ( ) = = More Problems 1. (2006 AMC 10B Problem 17) Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob s bag. Bob then randomly selects one ball from his bag and puts it into Alice s bag. What is the probability that after this process the contents of the two bags are the same? The contents of the two bags can only become the same if the ball Bob selects is the same color as the ball Alice selected. At the time of selection, Bob s bag has 6 balls, two of which have the correct color: one originally Bob s, and one originally Alice s. So the probability that Bob chooses one of them is 2 6 = You have a bag with 5 green balls, 2 yellow balls, and 6 red balls. You take balls out without putting them back in the bag, until there are no balls left. What s the probability that the third to last ball you draw is green or yellow? For any ball, considered individually, the probability is the same. One way to see this is to think of the drawing as choosing a random permutation of balls: for example, by laying them out in order, as you draw them. Then any individual ball has an equal chance to occupy the 7 third-to-last position. Out of 13 balls, 7 are green or yellow, so the probability is What s the probability that a randomly chosen integer between 1 and 1000 inclusive contains the digit 0? This is most easily done by counting integers with the digit 0 in them.

5 these equal and solving for p, we get p = (2007 AIME II Problem 10) Let S be a set with six elements. Let P be the set of all subsets of S. Subsets A and B of S, not necessarily distinct, are chosen independently and at random from P. The probability that B is contained in at least one of A or S \ A is m n r, where m, n, and r are positive integers, n is prime, and m and n are relatively prime. Find m + n + r. (The set S \ A is the set of all elements of S which are not in A.) The probability that B A is the same as the probability that B S \ A, since if A is equally likely to be any set in P, so is S \ A. So we begin by finding the former. We can think of A and B as being chosen by flipping fair coins, for each element x S, to see if x A and then to see if x B. We have B A if, for every x, either x A or x / B. This has a chance of happening for each x S, so overall we have Pr[B A] =. 4 6 We also have Pr[B S \ A] = 36. However, both can happen at the same time: if B =, 4 6 then B A and B S \ A for any A. This has a 1 chance of happening. 2 6 So the overall probability is = , giving a final answer of = You re playing a game that involves scoring points by collecting cubes of the same color. Every time you receive a cube, there is a 1/2 chance that it is white, a 1/3 chance that it is blue, and a 1/6 chance that it is a red cube. Your total score at the end of the game is the number of white cubes you have plus the product of the number of red and the number of blue cubes you have collected. For example, if you have 5 white cubes, 3 blue cubes, and 2 red cubes, your score would be = 11 points. What is the expected value of the number of points you will have, if you receive 10 cubes over the course of the game? (The expected value of a random value is the mean value, for example, the expected value of a die roll is 3.5.) Let W, B, and R be the number of white, blue, and red cubes drawn. We can separately compute the expected value of W and of BR. The expected value of W is easy: each cube has a 1 2 chance of contributing a point to W, so the average contribution of all ten cubes is = 5. To compute the expected value of BR, we think of this number as counting the ordered pairs (i, j) such that the i th cube is blue and the j th cube is red. There are 100 such pairs, but 10 of them are pairs (i, i), which can t possibly be the right combination of colors. Of the remaining 90, each pair has a expected value of = 5. 6 = 1 18 chance of being the right pair of colors, for an overall Putting these together, we conclude that the overall expected value is = 10.

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