Respuesta :

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

[tex]D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6[/tex]

Total outcomes

Combinations of 2 from a set of 52(number of playing cards). So

[tex]T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326[/tex]

What is the probability that they are both queens?

[tex]P = \frac{D}{T} = \frac{6}{1326} = 0.0045[/tex]

0.45% probability that they are both queens.

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