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Dear fellows. Yesterday I was trying to solve an interesting exercise of 
probabilities but I still do not understand the reasons of the solutions 
provided by the book. I will copy the exercise and my answers and then 
maybe someone of the list could help me with some advices to understand 
the solution (I'm traslating the exercise from Spanish, so please excuse 
any non-familiar term):

Three students participate in a card's game. They decide that the first 
person to play will be the one
who pick the major card in a 52 card's deck. They order the suits from 
lower than higher: clubs(?), diamonds (?), hearts (?) and spades (?).

a. If the card is thrown again into the card's deck after each student 
chose one, how many possibly configurations of the three options are?

Answer: 52*52*52 = 140608 (since any student can choose any card at any 
time)

b. How many configurations are in which each student selects a different 
card?

Answer: 52*51*50 = 132600

c. What is the probability that the three students select exactly the 
same card?

The answer is 0.00037, that corresponds to 52/140608, but I do not 
understand why 52 is on the numerator. Maybe the answer is correct but 
the form to calculate that is wrong... HELP REQUIRED

d. What is the probability that the three students choose different cards?

The answer is 0.943. This figure corresponds to: (52*51*50/140608), but 
I do not understand why the sample space is the one that corresponds to 
all arrangements possible when they select the same card. HELP REQUIRED

Thanks to all of you!
-- 

__________________________________________________________________

*Rodrigo Briceño*
Project Manager
Sanigest Internacional

+506  22-91-12-00 ext. 113  * Oficina* *Costa Rica*
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