If you are dealt 4 cards from a shuffled deck of 52 cards, find the probability of getting two queens and two kings.
1 answer:
<u>Given</u>:
If you are dealt 4 cards from a shuffled deck of 52 cards.
We need to determine the probability of getting two queens and two kings.
<u>Probability of getting two queens and two kings:</u>
The number of ways of getting two queens is
The number of ways of getting two kings is
Total number of cases is
The probability of getting two queens and two kings is given by
Substituting the values, we get;
Simplifying, we get;
Thus, the probability of getting two queens and two kings is 0.000133
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Answer:
(a) F(A, B, C) = AB'C' + A'BC' + A'B'C
(b) F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)
Step-by-step explanation:
(a) If F(A, B, C) = 1 iff exactly one of the coins is heads, then either
A is heads and the others are tails (AB'C')
B is heads and the others are tails (A'BC')
C is heads and the others are tails (A'B'C)
Hence, as a minterm expansion,
F(A, B, C) = AB'C' + A'BC' + A'B'C
(b) To get the corresponding maxterm expansion, we convert to binary.
The maxterm is the product of the complements.
Expanding,
F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)
We have 3⁴ = 81, so we can factorize this as a difference of squares twice:
Depending on the precise definition of "completely" in this context, you can go a bit further and factorize as yet another difference of squares:
And if you're working over the field of complex numbers, you can go even further. For instance,
But I think you'd be fine stopping at the first result,