Problem:
Jackie and Phil have two fair coins and a third coin that comes up heads with probability 74​. Jackie flips the three coins, and then Phil flips the three coins. Let nm​ be the probability that Jackie gets the same number of heads as Phil, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let p(h) be the probability that Jackie flips h heads. Then
​p(0)=(21​)2⋅73​=283​p(1)=2⋅41​⋅73​+41​⋅74​=145​p(2)=41​⋅73​+2⋅41​⋅74​=2811​, and p(3)=41​⋅74​=71​​
The probability that Jackie and Phil flip exactly the same number of heads is [p(0)]2+[p(1)]2+[p(2)]2+[p(3)]2=(283​)2+(145​)2+(2811​)2+(71​)2=392123​, and the requested sum is 123+392=515​.
The problems on this page are the property of the MAA's American Mathematics Competitions