mmsanotherstage2019.com->Probability-and-statistics-> SOLUTION: having trouble through this homework! Compute the probability of randomly drawing five cards from a deck and getting 3 Aces and 2 Kings. (Give her answer as a fraction.) var visible_logon_form_ = false;Log in or register.Username: Password: register in one easy step!.Reset her password if girlfriend forgot it."; return false; } "> log in On

Click below to view ALL problems on Probability-and-statisticsQuestion 1050727: having trouble v this homework! Compute the probability of randomly illustration five cards from a deck and also getting 3 Aces and 2 Kings. (Give her answer together a fraction.) price by Theo(11765) (Show Source): You deserve to put this equipment on your website! you attract 5 cards from the deck.the probability that you would acquire 3 aces and also 2 monarchs would be calculated as follows:the variety of ways you can acquire 3 aces the end of 4 aces is c(4,3) = 4.the number of ways girlfriend can obtain 2 majesties out the 4 majesties is c(4,2) = 6.the variety of ways friend can obtain 2 kings and 3 aces is 4 * 6 = 24.the variety of ways friend can obtain 5 cards out of a deck of 52 cards is c(52,5) = 2598960.the probability of gaining 3 aces and 2 emperors when you draw 5 cards native the deck is 24 / 2598960 = 9.234463016 * 10^-6.there is another method to analysis it.the answer must be the same.the probability of gaining 3 aces and also 2 monarchs would be:4/52 * 3/51 * 2/50 * 4/49 * 3/48 * 5! / (3! * 2!).use your calculator to gain 9.234463016 * 10^-6.the answers are the same.the two approaches are good.c(n,x) is same to n! / (x! * (n-x)!).this is the combination formula because that the variety of ways of acquiring x things out of n whereby order is no important.if you desire to get 3 aces out of 4 possible, the formula i do not care c(4,3) = 4! / (3! * 1!) = (4*3*2*1) / (3*2*1*1) = 4.if you want to get 2 monarchs out of 4 possible, the formula i do not care c(4,2) = 4! / (2! * 2!) = (4*3*2*1) / (2*1*2*1) = 6.the variety of possible combinations of 3 aces and also 2 queens becomes 4 * 6 = 24.the number of ways friend can obtain 5 cards, nevertheless of what lock are, the end of 52 cards is c(52,5) = 52! / (5! * 47!) = (52*51*50*49*48*47!) / (5*4*3*2*1*47!)/the 47! in the numerator and denominator publication out and also you room left v (52*51*50*49*48) / (5*4*3*2*1) i m sorry is same to 2598960.24/2598960 = 9.234463016 * 10^-6 as we calculated before.the second technique finds the probability that of gaining them in a specific order and then multiplies through the number of ways you have the right to arrange them.(4/52 * 3/51 * 2/50) is the probability of gaining the 3 aces.(4/49 * 4/48) is the probability of acquiring the 2 majesties once you"ve attracted the 3 aces.(4/52 * 3/51 * 2/50 * 4/49 * 4/48) is the probability of gaining the 3 aces and also the 4 queens in that certain order (first the 3 aces and also then the 2 kings).these 5 cards have the right to be i ordered it in 5! ways.however, 3 the them room the same and 2 the them space the same, therefore, the number of ways they deserve to be arranged becomes 5! / (3! * 2!).you room not distinguishing between the aces or the kings in anyway, such as suit, as such they space assumed to it is in the exact same cards.an ace is one ace nevertheless if it"s a spade or a heart.both methods are valid and also useful come know.i usually shot to settle these both means as a check against each method.if the outcomes are different, i try to discover out why.i expect you uncovered this helpful.