My Math Forum Probability problem.

 February 23rd, 2012, 09:16 PM #1 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Probability problem. Suppose that you are dealt a hand of 6 cards from a standard deck of 52 playing cards. a) What is the probability that you are dealt no jacks and exactly 1 diamond? b) What is the probability that you are dealt no more than 2 spades? c) What is the probability that you are dealt at least 2 sevens and at least 2 fives?
 February 24th, 2012, 01:30 AM #2 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Probability problem. a) $\frac{\binom{26}{5} \cdot 13}{\binom{52}{6}}$
February 24th, 2012, 06:40 AM   #3
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Re: Probability problem.

Hello, Ataloss!

Quote:
 Suppose that you are dealt a hand of 6 cards from a standard deck of 52 playing cards. b) What is the probability that you are dealt no more than 2 spades?

$\text{There are 13 Spades and 39 Others.}$

[color=beige]. . [/color]$\text{0 Spades: }\:{39\choose6}\text{ ways.}$

[color=beige]. . [/color]$\text{1 Spade: }\:{13\choose1}{39\choose5}\text{ ways.}$

[color=beige]. . [/color]$\text{2 Spades: }\:{13\choose2}{39\choose4}\text{ ways.}$

$\text{Therefore: }\:P(\text{at most 2 Spades}) \;=\;\frac{{39\choose6}\,+\,{13\choose1}{39\choose 5}\,+\,{13\choose2}{39\choose4}}{{52\choose6}}$

Edit: corrected my typo.

Quote:
 c) What is the probability that you are dealt at least two 7's and at least two 5's?

There are four 7's, four 5's, and 44 Others.
There are six case to consider.

$\begin{array}{ccc} &77,\,55,\,OO:\;{4\choose2}{4\choose2}{44\choos e2} \\ \\ \\

77,\,555,\,O:\;{4\choose2}{4\choose2}{44\choose1} && 777,\,55,\,O:\;{4\choose3}{4\choose2}{44\choose1} \\ \\ \\

77,\,5555:\:{4\choose2}{4\choose4} && 7777,\,55:\;{4\choose4}{4\choose2} \\ \\ \\

& 777,\,555:\:{4\choose3}{4\choose3} \end{array}$

I'll let you finish up . . .

February 24th, 2012, 08:41 AM   #4
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Re: Probability problem.

Quote:
 Originally Posted by soroban $\text{Therefore: }\:P(\text{at most 2 Spades}) \;=\;\frac{{39\choose6}\,+\,{39\choose1}{39\choose 5}\,+\,{13\choose2}{39\choose4}}{{52\choose6}}$
Curious, Soroban (or anyone): what does that end up to be as a single fraction?
Thanks.

February 24th, 2012, 04:40 PM   #5
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Re: Probability problem.

Quote:
Originally Posted by soroban
Hello, Ataloss!

Quote:
 Suppose that you are dealt a hand of 6 cards from a standard deck of 52 playing cards. b) What is the probability that you are dealt no more than 2 spades?

$\text{There are 13 Spades and 39 Others.}$

[color=beige]. . [/color]$\text{0 Spades: }\:{39\choose6}\text{ ways.}$

[color=beige]. . [/color]$\text{1 Spade: }\:{13\choose1}{39\choose5}\text{ ways.}$

[color=beige]. . [/color]$\text{2 Spades: }\:{13\choose2}{39\choose4}\text{ ways.}$

$\text{Therefore: }\:P(\text{at most 2 Spades}) \;=\;\frac{{39\choose6}\,+\,{39\choose1}{39\choose 5}\,+\,{13\choose2}{39\choose4}}{{52\choose6}}$

[quote:n6nvu1q1]c) What is the probability that you are dealt at least two 7's and at least two 5's?

There are four 7's, four 5's, and 44 Others.
There are six case to consider.

$\begin{array}{ccc} &77,\,55,\,OO:\;{4\choose2}{4\choose2}{44\choos e2} \\ \\ \\

77,\,555,\,O:\;{4\choose2}{4\choose2}{44\choose1} && 777,\,55,\,O:\;{4\choose3}{4\choose2}{44\choose1} \\ \\ \\

77,\,5555:\:{4\choose2}{4\choose4} && 7777,\,55:\;{4\choose4}{4\choose2} \\ \\ \\

& 777,\,555:\:{4\choose3}{4\choose3} \end{array}$

I'll let you finish up . . .

[/quote:n6nvu1q1]

this is really confusing, is it possible for you to write it in a different way?

February 24th, 2012, 07:23 PM   #6
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From: St. Augustine, FL., U.S.A.'s oldest city

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Re: Probability problem.

Quote:
Originally Posted by Denis
Quote:
 Originally Posted by soroban $\text{Therefore: }\:P(\text{at most 2 Spades}) \;=\;\frac{{39\choose6}\,+\,{39\choose1}{39\choose 5}\,+\,{13\choose2}{39\choose4}}{{52\choose6}}$
Curious, Soroban (or anyone): what does that end up to be as a single fraction?
Thanks.
Methinks soroban simply mistyped and meant:

$\text{Therefore: }\:P(\text{at most 2 Spades}) \;=\;\frac{{39\choose6}\,+\,{13\choose1}{39\choose 5}\,+\,{13\choose2}{39\choose4}}{{52\choose6}}=\fr ac{660117}{783020}\approx
0.843039769099129013307450639830400245204464764629 2559...$

Quote:
 Originally Posted by halloweengrl23 ... this is really confusing, is it possible for you to write it in a different way?
That's a first!

You will be hard pressed to find anyone whose explanations are more clear and thorough.

February 24th, 2012, 10:31 PM   #7
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Re: Probability problem.

$\text{Therefore: }\:P(\text{at most 2 Spades}) \;=\;\frac{{39\choose6}\,+\,{13\choose1}{39\choose 5}\,+\,{13\choose2}{39\choose4}}{{52\choose6}}$[/size]

Edit: corrected my typo.

Quote:
 c) What is the probability that you are dealt at least two 7's and at least two 5's?

There are four 7's, four 5's, and 44 Others.
There are six case to consider.

$\begin{array}{ccc} &77,\,55,\,OO:\;{4\choose2}{4\choose2}{44\choos e2} \\ \\ \\

77,\,555,\,O:\;{4\choose2}{4\choose2}{44\choose1} && 777,\,55,\,O:\;{4\choose3}{4\choose2}{44\choose1} \\ \\ \\

77,\,5555:\:{4\choose2}{4\choose4} && 7777,\,55:\;{4\choose4}{4\choose2} \\ \\ \\

& 777,\,555:\:{4\choose3}{4\choose3} \end{array}$

I'll let you finish up . . .

[/quote]

How do you solve this? I am not familair with this level of probability.

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