Probability: Marbles, Coins, Cards
Date: 05/24/97 at 00:00:44 From: Jean-Pierre Berard Subject: Help wanted Hello, I want to write a little program to calculate permutations, combinations, variations with repetitions, and so on. I thought that writing the formulas down would make the reasoning come clear. Right away, I was confronted with my own shortcomings in this area of math! In an effort to overcome this, could you please explain the following problems to me? 1) A bag contains 10 marbles, which are identical except for their color. There are 4 blue marbles and 6 red marbles. If marbles are drawn at random from the bag, what is the probability of drawing the following colored marbles in the order indicated: red, blue, blue, red. 2) The directors of a company can not decide who should be chairman and minute secretary. If there are 20 directors. a) How many different selections are possible?. b) What is the probability that any two given directors will become chairman and minute secretary?. c) What is the probability that any two given directors will become chairman or minute secretary?. 3) A card player draws 5 cards from a full deck of 52 cards. a) What is the probability that their hand of cards contains an ace of spades, a 6 of hearts, a 5 of clubs, a king of hearts and a 9 of diamonds? b) What is the probability that their hand of cards contains an ace of spades, a 6 of hearts, a 5 of clubs, a king of hearts and any diamond? 4) In a coin tossing experiment, an unbiased coin is flipped 8 times. What is the probability that 5 heads will result?. Many thanks, Jean-Pierre Berard
Date: 05/24/97 at 07:55:49 From: Doctor Mitteldorf Subject: Re: Help wanted Dear Jean-Pierre, I'm not sure how you want to program these. One very powerful and general method is called "Monte Carlo". You use the computer's number crunching capability, and you just simulate the entire experiment perhaps a million times, keeping track of how many times the computer achieves a successful outcome by whatever criterion you have. This method is a bit inelegant and slow, but useful in the many circumstances when there is no obvious formula or reasoning that will make the problem simpler. The answer is approximate, and gets better the more trials you simulate. You can do a Monte Carlo simulation of any of the 4 problems you've described, but they are also simple enough that they can all be done with analytic methods. I don't think you need a computer. 1) The probability of drawing the first red is 4/10, which leaves 3 red and 6 blue. The probability of drawing the second blue is 6/9, which leaves 3 red and 5 blue. The probability of drawing the third blue is 5/8 which leaves 3 red and 4 blue. The probability of drawing the fourth red is 3/7 (tt doesn't matter what this leaves.) The probability of the entire sequence is just the product of the probability of each individual drawing: 4/10 * 6/9 * 5/8 * 3/7 = 360/5040 = 1/14 2a) There are 20 possibilities for chairman, and having chosen chairman there are 19 remaining possibilities for secretary. b) The probability of any given combination of two people for chair and secretary is 1/380. c) I'm not sure what you mean by this question. Do you mean the probability that either Andrew will be chairman OR Bill will be secretary? Out of the 380 possibilities, there are 19 in which Andrew is chair, and 19 in which Bill is secretary, but these two sets overlap by 1, in which both conditions are met. The probability that either condition is met is thus 37/380. 3a) The probability of drawing these 5 cards in order is 1 in (52*51*50*49*48). But they don't have to be drawn in that particular order. How many different orders are there? This is the number of ways to order 5 cards, which is 5! or 5*4*3*2*1.So the probability of having the given hand is: (5*4*3*2)/(52*51*50*49*48) This is just the inverse of the number called "52 choose 5" and sometimes written as a large pair of parentheses enclosing a 52 and a 5 directly underneath it. The formula for "n choose r" is n!/((n-r)!*r!) 3b) This proability is just 13 times as large as (a). 4) If you count as a separate case each sequence (e.g. head-tail-head- head-head-tail-tail-head) then the total number of possible outcomes for 8 tosses is 2^8. This comes from the fact that each separate toss has 2 outcomes, and there are 8 tosses. You want to know, of these 2^8, how many have 5 heads and 3 tails? The answer is "8 choose 5" or (8*7*6) / (3*2*1). This is because there are 8! ways of arranging 8 coins if they are all different, but they're not all different: 5 of them are heads and indistinguishable, so you've overcounted by the orderings of 5 things, a factor of 5!. Also, 3 of them are tails and indistinguishable, so you've overcounted by a factor 3! The probability you want is "8 choose 5" divided by 2^8. -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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