Date: 06/12/97 at 21:27:46 From: Anonymous Subject: Math-probability What is the probability that you will pick two aces in a row out of a 52 card deck of playing cards? I have asked my parents, my teacher, my class and have even searched the entire Web for the answer. I don't understand how you would write the answer out. I have no area what I need help with specifically - I need to know the whole answer. It seems like the hardest question I have ever heard. I am hoping you can help me figure it out! Thanks, Anonymous
Date: 06/13/97 at 05:38:10 From: Doctor Anthony Subject: Re: Math-probability In fact the answer can be written down in a few seconds if you are familiar with the topic in mathematics known as permutations and combinations. I will give the answer in the notation of combinations, and then consider a more basic approach. You have to select 2 aces from the 4 available, and this can be done in 4_C_2 ("four choose two) ways = 6 ways. However the number of ways for selecting any two cards from 52 is 52_C_2 = 1326 ways. So the probability that you will select two aces when you remove two cards from the pack is given by: 4_C_2 6 1 -------- = ---- = ----- 52_C_2 1326 221 4 x 3 The expression 4_C_2 is calculated from ----- = 6 1 x 2 52 x 51 So 52_C_2 = --------- = 1326 1 x 2 The more complicated the problem, the more the use of combinatorial methods becomes necessary. However, the problem you asked about can be done without the use of combinations. If you have a pack of 52 cards and you draw one card out, the probability that it will be an ace is 4/52 because there are four aces in a deck of 52 cards. Now you will be down to 51 cards, and of these 3 will be aces, so if you draw a second card the probability that this one is also an ace is 3/51 (since there are only three aces and a total of 51 cards left). To get the probability that both events will happen, i.e. an ace on the first draw and an ace on the second draw, we must multiply the probabilities. 4 3 1 Probability of two aces = ---- x --- = --- 52 51 221 This is same result we got before. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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