Dear Rodrigo,
Below is my solution to points c and d.
Rodrigo Briceņo wrote:
> 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
The first student, say A, selects one card (any of the 52, it does not
matter which one). The probability that the second, say B, selects the
same card is 1/52. Now, the probability that the third student, C,
selects the same card is 1/52 times the probability that B has selected
the same card as A, ie, 1/52. Thus, the result is 1/52 * 1/52 = 0.00037.
>
> 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
>
The reasoning is the same as above. The first student, say A, selects
one card (any of the 52, it does not matter which one). The probability
that B selects a different card is 51/52, ie, that he selects all but
the chosen card by A. The probability that C selects a different card is
50/52 (ie, the probability to select all but the two already chosen
cards) times the probability that A and B have selected different cards
(51/52). Therefore the result is 50/52 * 51/52 = 0.943.
> Thanks to all of you!
Kind regards,
Cristian Gatu
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