Problem:
Two three-letter strings, aaa and bbb, are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 31β chance of being received incorrectly, as an a when it should have been a b, or as a b when it should have been an a. However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let Saβ be the threeletter string received when aaa is transmitted and let Sbβ be the three-letter string received when bbb is transmitted. Let p be the probability that Saβ comes before Sbβ in alphabetical order. When p is written as a fraction in lowest terms, what is its numerator?
Solution:
Let Saβ, the three-letter string received when aaa is transmitted, be x1βx2βx3β and let Sbβ be y1βy2βy3β, where each of the xkβ,ykβ is an a or a b. It will be convenient to introduce the symbol βΊ to denote that one string of letters precedes another in alphabetical order. (Thus, if S1β and S2β are two strings of letters, then S1ββΊS2β is to be read "S1β precedes S2β alphabetically.") We will find the probability that SaββΊSbβ. Since the reception of any one letter is independent of that of any of the other letters, we have
Prob(SaββΊSbβ)β=Prob(x1βx2βx3ββΊy1βy2βy3β)=Prob(x1ββΊy1β)+Prob(x1β=y1β and x2ββΊy2β)+Prob(x1β=y1β and x2β=y2β and x3ββΊy3β)=Prob(x1ββΊy1β)+Prob(x1β=y1β)β
Prob(x2ββΊy2β)+Prob(x1β=y1β)β
Prob(x2β=y2β)β
Prob(x3ββΊy3β).(β)β
Now x1ββΊy1β is true if and only if x1β=a and y1β=b; that is, if and only if these leading letters were received correctly. Since for each letter there is a 32β probability that it was received correctly, we conclude that
Prob(x1ββΊy1β)=32ββ
32β=94β.
Similarly, Prob(x2ββΊy2β)=Prob(x3ββΊy3β)=94β. The relation x1β=y1β is true if and only if one of these letters was received correctly and the other was received incorrectly. Thus
Prob(x1β=y1β)=Prob(x1β=y1β=a)+Prob(x1β=y1β=b)=32ββ
31β+31ββ
32β=94β.
Identical reasoning shows that Prob(x2β=y2β)=94β also. Substituting these probabilities in (β) we have
Prob(SaββΊSbβ)=94β+(94β)2+(94β)3=729532β
The desired numerator is 532β.
The problems on this page are the property of the MAA's American Mathematics Competitions