one more probability?

Question

From a standard deck of 52 cards, two cards are dealt without replacement. What is the probability that both are aces OR both are black?

Please setup in terms of nCr or combination form.

Where I get confused is "or both are black" because it overlaps the two possible black aces so don't i have to subtract or something like that?

Answer 1

Yes, but there's only one combination of both black aces.

So you have (4C2 + 26C2 - 1)/52C2 or whatever method you're using.

Answer 2

Let A be the event that two aces are picked , B be the event that two black cards are picked, and C be the event that two black aces are picked. (Note that event C is the intersection of A and B.)

P(A) = (4/52)*(3/51)
P(B) = (26/52)*(25/52)
P(C) = (2/52)*(1/51)

As you noted, A and B are not independent. Therefore P(A) + P(B) over counts the event you're interested in by exactly P(C). (Drawing a Venn diagram of this can be helpful.)

Thus, the probability of A or B occurring (A union B) = P(A)+P(B)-P(C).

Answer 3

You're after the probability P(A) + P(B) - P(AB)
P(A) = (4/52)(3/51)
P(B) = (26/52)(25/51)
P(AB) = (2/51)(1/51)
P = (4*3 + 26*25 - 2)/(52*51)

Answer 4

the prob they are both aces is (4c2)/(52c2)
the prob they are both black is (26c2)/(52c2)

I guess the probability they are both black or aces i guess would just mean adding them together. Unless you need to subtract the possibilty they are both black and aces with would be 1/(52c2)