Difference between revisions of "2007 iTest Problems/Problem 4"
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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 | + | 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== | ||
{{iTest box|year=2007|num-b=3|num-a=5}} | {{iTest box|year=2007|num-b=3|num-a=5}} | ||
[[Category:Introductory Combinatorics Problems]] | [[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.
Solution
The number of arrangements of the group of letters HHTT is 
, so the probability that two of the flips are heads is 
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 3 |
Followed by: Problem 5 | |
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