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=283p(1)=2⋅41⋅73+41⋅74=145p(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