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Difference between revisions of "2006 Alabama ARML TST Problems/Problem 3"

(New page: ==Problem== River draws four cards from a standard 52 card deck of playing cards. Exactly 3 of them are 2’s. Find the probability River drew exactly one spade and one club from the deck....)
 
(→‎Case 2: Both)
 
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==Solution==
 
==Solution==
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The probability is equal two the number of successful outcomes(<math>S</math>) divided by the number of outcomes(<math>N</math>). <math>N=4\cdot 48</math>, from the 4 ways to choose the 2's and the 48 ways to choose the other card. Now we find <math>S</math>. From the three 2's, there must be at least one spade or club.
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The probability is equal to the number of successful outcomes(<math>S</math>) divided by the number of outcomes(<math>N</math>). <math>N=4\cdot 48</math>, from the 4 ways to choose the 2's and the 48 ways to choose the other card. Now we find <math>S</math>. From the three 2's, there must be at least one spade or club.
  
 
===Case 1: One but not the other===
 
===Case 1: One but not the other===
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===Case 2: Both===
 
===Case 2: Both===
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There are two ways to choose the third 2, and then we must choose a heart or a diamond, which there are 25 of.
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There are two ways to choose the third 2, and then we must choose a heart or a diamond, which there are 24 of (We can't choose another 2 of Hearts or Diamonds because that would give us four 2's instead of three).
  
 
===Answer===
 
===Answer===
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Therefore, <math>S=24+2\cdot 25=74</math>. Thus the probability of one spade and one club is <math>\boxed{\dfrac{37}{96}}</math>
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Therefore, <math>S=24+2\cdot 24=72</math>. Thus the probability of one spade and one club is <math>\boxed{\dfrac{3}{8}}</math>
  
 
==See also==
 
==See also==
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{{ARML box|year=2006|state=Alabama|num-b=2|num-a=4}}

Latest revision as of 19:43, 12 April 2012

Problem

River draws four cards from a standard 52 card deck of playing cards. Exactly 3 of them are 2’s. Find the probability River drew exactly one spade and one club from the deck.

Solution

The probability is equal to the number of successful outcomes($S$) divided by the number of outcomes($N$). $N=4\cdot 48$, from the 4 ways to choose the 2's and the 48 ways to choose the other card. Now we find $S$. From the three 2's, there must be at least one spade or club.

Case 1: One but not the other

Whether it's a spade or a club in the 2's, the probability is the same, so we must multiply by two. Now the number of ways to choose a spade but not a club is 12, since after we choose the 3 2's, we must choose a club that is not a 2. $12\cdot 2=24$.

Case 2: Both

There are two ways to choose the third 2, and then we must choose a heart or a diamond, which there are 24 of (We can't choose another 2 of Hearts or Diamonds because that would give us four 2's instead of three).

Answer

Therefore, $S=24+2\cdot 24=72$. Thus the probability of one spade and one club is $\boxed{\dfrac{3}{8}}$

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 2 		Followed by:
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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