My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Thanks Tree1Thanks
  • 1 Post By Prove It
Reply
 
LinkBack Thread Tools Display Modes
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
taylor_1989_2012 is offline  
 
October 30th, 2016, 04:37 PM   #2
Member
 
Joined: Oct 2016
From: Melbourne

Posts: 77
Thanks: 35

Quote:
Originally Posted by taylor_1989_2012 View Post
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.

$\displaystyle \begin{align*} 4 \cdot \frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{9}{48} &= 4 \cdot \frac{1}{4} \cdot \frac{4}{17} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{3}{16} \\ &= 1 \cdot \frac{1}{17} \cdot \frac{11}{5} \cdot \frac{1}{49} \cdot \frac{3}{4} \\ &= \frac{33}{16\,660} \end{align*}$
Thanks from topsquark
Prove It is offline  
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)?
taylor_1989_2012 is offline  
October 30th, 2016, 07:11 PM   #4
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,588
Thanks: 1038

Quote:
Originally Posted by taylor_1989_2012 View Post
a) All are of the same suit?
12/51 * 11/50 * 10/49 * 9/48 = 13199/66640 = .198

No need to use 13/52...
Denis is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
cards, deck



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
A deck of cards pedro malafaya Elementary Math 3 May 27th, 2013 07:56 AM
Probability with Deck of Cards flyingL123 Advanced Statistics 1 October 16th, 2011 10:05 PM
Deck of Cards Probability carolinegustin Probability and Statistics 3 March 27th, 2011 07:28 PM
Fair deck of 52 Cards wowbringer Algebra 1 March 17th, 2011 09:05 PM
Possibilities with a standard deck of cards? Krimi Algebra 3 January 19th, 2008 07:52 PM





Copyright © 2019 My Math Forum. All rights reserved.