Difference between revisions of "2007 iTest Problems/Problem 4"

(New page: ==Problem== Star flips a quarter four times. Find the probability that the quarter lands heads exactly twice. <math>\mathrm{(A)}\,\frac{1}{8}\quad\mathrm{(B)}\,\frac{3}{16}\quad\mathrm{(C...)
 
(Solution)
 
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==Solution==
 
==Solution==
The number of arrangements of the group of letters HHTT is <math>\frac{4!}{2!*2!}=6</math>, so the probability that two of the flips are heads is <math>\frac{6}{2^4}=\frac{3}{8} \Rightarrow D</math>
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The number of arrangements of the group of letters HHTT is <math>\frac{4!}{2!*2!}=6</math>, so the probability that two of the flips are heads is <math>\frac{6}{2^4}=\frac{3}{8} \Rightarrow C</math>
  
 
==See Also==
 
==See Also==
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{{iTest box|year=2007|num-b=3|num-a=5}}
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[[Category:Introductory Combinatorics Problems]]

Latest revision as of 18:59, 21 May 2008

Problem

Star flips a quarter four times. Find the probability that the quarter lands heads exactly twice.

$\mathrm{(A)}\,\frac{1}{8}\quad\mathrm{(B)}\,\frac{3}{16}\quad\mathrm{(C)}\,\frac{3}{8}\quad\mathrm{(D)}\,\frac{1}{2}$

Solution

The number of arrangements of the group of letters HHTT is $\frac{4!}{2!*2!}=6$, so the probability that two of the flips are heads is $\frac{6}{2^4}=\frac{3}{8} \Rightarrow C$

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 3
Followed by:
Problem 5
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