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 October 30th, 2016, 04:28 PM #1 Senior Member   Joined: Sep 2012 Posts: 201 Thanks: 1 Deck of cards Hi again. I have been on this question for sometime and really cant seem to solve it any help would be appreciated. Five cards are dealt from a standard shuffled deck. What is the probability that: a) All are of the same suit? b) There is at least one card from each of the four suits? Method for a) If I were to choose say Hearts twice and then I would get the following: $\displaystyle \frac{13}{52}*\frac{12}{51}*\frac{11}{50}*\frac{10 }{49}*\frac{9}{48}$ I then times by 4 for each suit and the by 100 and got approx $\displaystyle 0.2%$ I the checked this method by comparing to combinations: $\displaystyle \frac{13C5}{52C5}$ This would be for me picking say two hears again I then multiplyed this by 4 to geth the same ans Now be I am slightly confused, so buy using combinations I got these results# For 2 cards of the same set I calculated the probability like so: $\displaystyle \frac{13C2*13^3*4}{52C5}$ For 3 cards $\displaystyle \frac{13C3*13^2*4}{52C5}$ For 4 cards $\displaystyle \frac{13C4*13*4}{52C5}$ And I already calculated 5 cards, but shouldn't my probability add up to one, which is why I think I am wrong in the method I have chosen. Could someone please give me a point in the right direction please
October 30th, 2016, 04:37 PM   #2
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 Originally Posted by taylor_1989_2012 Hi again. I have been on this question for sometime and really cant seem to solve it any help would be appreciated. Five cards are dealt from a standard shuffled deck. What is the probability that: a) All are of the same suit? b) There is at least one card from each of the four suits? Method for a) If I were to choose say Hearts twice and then I would get the following: $\displaystyle \frac{13}{52}*\frac{12}{51}*\frac{11}{50}*\frac{10 }{49}*\frac{9}{48}$ I then times by 4 for each suit and the by 100 and got approx $\displaystyle 0.2%$ I the checked this method by comparing to combinations: $\displaystyle \frac{13C5}{52C5}$ This would be for me picking say two hears again I then multiplyed this by 4 to geth the same ans Now be I am slightly confused, so buy using combinations I got these results# For 2 cards of the same set I calculated the probability like so: $\displaystyle \frac{13C2*13^3*4}{52C5}$ For 3 cards $\displaystyle \frac{13C3*13^2*4}{52cdC5}$ For 4 cards $\displaystyle \frac{13C4*13*4}{52C5}$ And I already calculated 5 cards, but shouldn't my probability add up to one, which is why I think I am wrong in the method I have chosen. Could someone please give me a point in the right direction please
1. I agree with your reasoning, but DO NOT APPROXIMATE! Just simplify your fraction.

\displaystyle \begin{align*} 4 \cdot \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{9}{48} &= 4 \cdot \frac{1}{4} \cdot \frac{4}{17} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{3}{16} \\ &= 1 \cdot \frac{1}{17} \cdot \frac{11}{5} \cdot \frac{1}{49} \cdot \frac{3}{4} \\ &= \frac{33}{16\,660} \end{align*}

 October 30th, 2016, 04:39 PM #3 Senior Member   Joined: Sep 2012 Posts: 201 Thanks: 1 Thank you for the quick response, may I ask what you think of my ans to b)?
October 30th, 2016, 07:11 PM   #4
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Quote:
 Originally Posted by taylor_1989_2012 a) All are of the same suit?
12/51 * 11/50 * 10/49 * 9/48 = 13199/66640 = .198

No need to use 13/52...

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