My Math Forum Probability of drawing a set of cards given several different drawing parameters

 October 16th, 2014, 02:26 PM #2 Newbie   Joined: Apr 2014 From: México Posts: 10 Thanks: 0 Math Focus: Probability Ho Kinead, at least for the first question I think you are right. When I have doubts about the solution of a particular problem I use to simulate the problem. As you will see, the aproximate solution from the simulation is very close to your answer. I think you can do something similar for the other questions. Good look. The following code is written in R (open source), I hope it is useful for you. Code: > library(prob) > deck <- cards() > > #simulate the experiment of draw (random) 7 cards from deck without replacenment > draw <- function(ncards = 7, from = deck){ + index <- sample(1:nrow(from), size = ncards,replace = FALSE) + deck[index,] + } > > > #replicate experimento draw(7,deck) > experiment <- replicate(n = 1000000, expr = draw()) > S <- data.frame(matrix(unlist(experiment),nrow=1000000,byrow=TRUE)) > S <- probspace(S) #add approximate probs for each result > #some results are > head(S) X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 probs 1 9 8 3 8 K K J Club Diamond Diamond Heart Diamond Heart Diamond 1e-06 2 2 Q 4 9 9 7 A Heart Heart Diamond Diamond Club Diamond Club 1e-06 3 8 3 K 3 7 J 9 Diamond Club Heart Spade Diamond Diamond Diamond 1e-06 4 4 K 2 9 3 Q Q Diamond Diamond Spade Heart Club Diamond Spade 1e-06 5 7 6 3 6 7 3 6 Club Diamond Diamond Club Heart Spade Heart 1e-06 6 Q 8 7 8 K 10 9 Spade Heart Diamond Spade Club Club Spade 1e-06 > #approximate probability for al least 1 ace since we took a 7 cards without replacenment > A <- S[which(apply(S,1,function(row) sum(row[1:7] == "A")) >= 1),] > Prob(A) [1] 0.450524 > #Your answer > 1 - choose(48,7)/choose(52,7) [1] 0.4496445
October 16th, 2014, 10:07 PM   #3
Math Team

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From: Lexington, MA

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Quote:
 Find the probability drawing at least one Ace from a deck of cards, given 7 draws from the deck (without replacement). 1. I believe the equation is: P(at least one Ace) = 1 - C(48,7)/C(52,7) (is this correct?)$\;\; \color{blue}{\text{Right!}}$ 2. I also think that if you were replacing the cards to the deck each time, the equation would be: P(at least one Ace) = 1 - (48/52)^7 (again, is this correct?)$\;\;\color{blue}{\text{Yes!}}$
Quote:
 Now it starts getting more difficult. Find the probability of drawing at least 2 Aces in the same 7 draws?
$P(\text{no Aces}) \:=\:\dfrac{_{48}C_7}{_{52}C_7} \qquad P(\text{one Ace}) \:=\:\dfrac{4(_{48}C_6)}{_{52}C_7}$

$P(\text{at least 2 Aces}) \:=\:1 - P(\text{no Aces}) - P(\text{one Ace})$

 Tags cards, drawing, parameters, probability, set

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### four cards are drawn from 52 cards with replacement find the probability of getting atleast 3aces

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