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 Probability and Statistics Basic Probability and Statistics Math Forum

 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... Tags cards, deck Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pedro malafaya Elementary Math 3 May 27th, 2013 07:56 AM flyingL123 Advanced Statistics 1 October 16th, 2011 10:05 PM carolinegustin Probability and Statistics 3 March 27th, 2011 07:28 PM wowbringer Algebra 1 March 17th, 2011 09:05 PM Krimi Algebra 3 January 19th, 2008 07:52 PM

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