SOLUTION: A sign reads “ALABAMA”, a drunk man removed 4 letters and then returned back
randomly into the four empty spaces again. What is the probability that the sign
still reads “A
Algebra.Com
Question 1176649: A sign reads “ALABAMA”, a drunk man removed 4 letters and then returned back
randomly into the four empty spaces again. What is the probability that the sign
still reads “ALABAMA”?
Found 2 solutions by Solver92311, Edwin McCravy:
Answer by Solver92311(821) (Show Source): You can put this solution on YOUR website!
Depends. Which four letters were removed. If he took the four As, then the probability of the sign reading ALABAMA when he put the letters back is 100%. But if he took four different letters, there are 24 ways to arrange four things only one of which would be correct. Different answer if he took two or three As.
John

My calculator said it, I believe it, that settles it
From
I > Ø
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!
The above is wrong.
A sign reads “ALABAMA”, a drunk man removed 4 letters and then returned back
randomly into the four empty spaces again. What is the probability that the
sign still reads “ALABAMA”?
The order he places them on the sign doesn't matter, for if he picked the 4
A's, the sign will read "ALABAMA" regardless of where he puts them, and if he
didn't pick them, it won't spell "ALABAMA", no matter where he puts them.
So there is only 1 way he can succeed, and that's by choosing the 4 A's.
The number of ways he can pick any 4 of the 7 letters is "7 letters choose 4" or
C(7,4) = 35.
So the probability is 1 way out of 35 or 1/35.
Edwin
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