SOLUTION: Joey prepares 3 cards for his 3 girlfriends. He addresses 3 corresponding envelopes. A brown-out suddenly occurred and he hurriedly placed the cards in the envelope at random. What

Algebra ->  Probability-and-statistics -> SOLUTION: Joey prepares 3 cards for his 3 girlfriends. He addresses 3 corresponding envelopes. A brown-out suddenly occurred and he hurriedly placed the cards in the envelope at random. What      Log On


   



Question 849374: Joey prepares 3 cards for his 3 girlfriends. He addresses 3 corresponding envelopes. A brown-out suddenly occurred and he hurriedly placed the cards in the envelope at random. What is the probability that
a. Each card is sent to its proper addressee?
b. No card is sent to the proper addressee?

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Joey prepares 3 cards for his 3 girlfriends. He addresses 3 corresponding
envelopes. A brown-out suddenly occurred and he hurriedly placed the cards in
the envelope at random. What is the probability that
a. Each card is sent to its proper addressee?
1 way out of 3! or 1/3! or 1/6

b. No card is sent to the proper addressee?
That's 1 out of 3 sub-factorial.  But if you haven't studied sub-factorials,
then we have to list the ways they could have all gone wrong. Girlfriend 1
could have been sent card 2 or 3, and in either case there is only one
way the other one could have gone wrong.

gf       1  2  3
card     2  3  1

or

gf       1  2  3
card     3  1  2

[Actually 3 sub-factorial = 3!(1/0!-1/1!+1/2!-1/3!) = 6(1/1-1/1+1/2-1/6) =
6(1/2-1/6) = 6(3/6-1/6) = 6(2/6) = 2]

So there are only 2 ways the cards could have all gone wrong: 

2/3! = 2/6 = 1/3

Edwin