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!
--
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*Rodrigo Briceño*
Project Manager
Sanigest Internacional
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