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You are told that of the four cards face down on the table, two are red and two are black. If you guess all four at random, what is the probability that you get 0, 2, 4 right ?

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You are told that of the four cards face down on the table, two are red and two are black. If you guess all four at random, what is the probability that you get 0, 2, 4 right ?

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CarlB

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I don't like this problem because it seems to me that it is worded in a manner that would allow it to have more than one solution.Alexsandro said:You are told that of the four cards face down on the table, two are red and two are black. If you guess all four at random, what is the probability that you get 0, 2, 4 right ?

(a) Suppose that by "guess all four at random" it means that you will guess each one independently with 50% red and 50% black. In this case, the fact that 2 of the cards are red and 2 are black doesn't matter. The probabilities for 0, 2 and 4 come from the binomial distribution and are

[tex](.5)^4\left(\begin{array}{c}4 \\ 0\end{array}\right)=\frac{1}{16}[/tex]

[tex](.5)^4\left(\begin{array}{c}4 \\ 2\end{array}\right)=\frac{6}{16}[/tex]

[tex](.5)^4\left(\begin{array}{c}4 \\ 4\end{array}\right)=\frac{1}{16}[/tex]

(b) Suppose that "guess all 4 at random" means to pick 2 and guess red, and guess black for the other two. Then your guesses are not independent. In this case, you may as well assume that you guess the first two red and the second two black. The probability that you get 0, 2 and 4 right are then:

[tex]\frac{2}{4}\frac{1}{3} = \frac{1}{6}[/tex]

[tex]\frac{4}{4}\frac{2}{3} = \frac{2}{3}[/tex]

[tex]\frac{2}{4}\frac{1}{3} = \frac{1}{6}[/tex]

If I were forced to grade this problem, I'd have to mark either of the above correct.

Carl

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