Question 1082753: In a particular card game, players receive five cards from a standard deck of 52 cards (consisting of the standard four suits, and 13 cards of each suit). Determine each of the following probabilities:
a) The probability of 2 or more red cards using a hypergeometric distribution
b) The probability of 2 or more red cards using a normal approximation. Be sure to check that it is appropriate to use an approximation.
Answer by natolino_2017(77)   (Show Source):
You can put this solution on YOUR website! a) P(X=a) = (dCa)((N-d)C(n-a))/(NCn)
in this case: N = 52 (number of total cards)
n = 5 (number of picked cards)
d = 26 (number of working cards)
According to the model, it's easier using the complement of the asked probability.
P() = 1 - (P(x=1) + P(x=0)
1 - (26C1)(26C4)/(52C5) - (26C2)(26C3)/(52C5) = 5,251/9,996 = 52.531%
b) First Not that the E(x) = nd/N = 5/2 ( So the expected number of red cards is between of two and three)
and V(x) = (N-n)(N-d)(nd)/(N^2(N-1)) = 235/204 (So the variation expected is aproximated one between every experiment).
As N = 52 > 30 it's fair to use a Normal aproximation.
y is a Normal with u= 2.5 and sigma square = 235/204
P(y>=2) using standardization P(Z >= (2-2.5)/(sqrt(235/204))
P(Z >= -0.46585) using symmetry.
0.5 + P(0 <= z <= 0.46585) Extrapolating
0.5 + 0.179 = 0.679 = 67.9%
which is different that the exact value, but the error is less than 30%.
If we were using several deck of cards the error would minor.
@natolino_
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