Probability of Poker Hands and Conditional Probability

Problem #1

A poker hand is defined as drawing five cards at random without replacement from a deck of 52 playing cards. Find the probability of the following power hands:
a) Four of a kind (four cards of equal face value and one of a different value).
b) Full house (one pair and one triple cars with equal face value).
c) Three of a kind (three equal face values plus two cards of different values).
d) Two pairs (two pairs of equal face value plus one card of different value).
e) One pair (one pair of equal face value plus three cards of different values).

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I know the probability of drawing five unordered cards at random without replacement is [sub:d8tev5w7]52[/sub:d8tev5w7]C[sub:d8tev5w7]5[/sub:d8tev5w7] = 2,598,960. That's what the example has. However, I'm having trouble calculating the probability of the four of a kind. If four of a kind has the same face value, then if I divide the 52 cards by its suite, I get 13 of each. In each of the suite, there's one card I'm looking for, except for the last one. The last one, there's two since I want to get one card of a different face value. So it would be [sub:d8tev5w7]13[/sub:d8tev5w7]C[sub:d8tev5w7]1[/sub:d8tev5w7]x [sub:d8tev5w7]13[/sub:d8tev5w7]C[sub:d8tev5w7]1[/sub:d8tev5w7]x [sub:d8tev5w7]13[/sub:d8tev5w7]C[sub:d8tev5w7]1[/sub:d8tev5w7]x [sub:d8tev5w7]13[/sub:d8tev5w7]C[sub:d8tev5w7]2[/sub:d8tev5w7]. But that gives me 171,366. Dividing it by 2,598,960, I got 0.05936375 which is not the answer. The answer is 0.00024.

Would anyone kindly help me with this one? I just need to know how to answer part a.

Problem #2:

An urn contains four colored balls: two orange and two blue. Two balls are selected at random without replacement, and you are told that at least one of them is orange. What is the probability that the other is also orange.

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To have the other ball orange, I concluded that both ball are orange. So the probability of the first drawn is 2/4. The probability that the second ball drawn is orange after the first one drawn is orange is 1/3. Using this rule, P(ANB) = P(B|A)P(A) = I have 1/6. But the answer is 1/5.

Problem #2:

A drawer contain four black, six brown, and eight olive socks. Two socks are selected at random from the drawer.
a) Computer the probability that both socks are the same color.
b) Computer the probability that both socks are olive if it is known that they are the same color.

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a) I was able to figure out the answer is 49/153.
b) I'm not sure how to figure out this one. From the words alone, it seems that I can use P(ANB) = P(B|A)P(A)
P(A) = The probability that both socks are the same color.
P(B) = he probability that both socks are olive.
If it is true I can use that, how can I calculate P(B|A)? Can I set 8/18 = x/153. P(B|A) = x/153? Actually I calculated that but it still didn't give me the right answer. I have x = 68.

Thank you in advance for your help!

Four of a kind:

There are 13 numbered cards and 48 left to choose from after they are drawn. So, \(\displaystyle 13\cdot 48=624\)

\(\displaystyle \frac{624}{C(52,5)}=\frac{1}{4165}\approx .00024\)

If you want to see it in combo format, then we are choosing 4 cards from 4 that are alike, Then 1 set from the 13 and 1 from the remaining 48.

\(\displaystyle \frac{C(4,4)\cdot C(13,1)\cdot C(48,1)}{C(52,5)}\)

Thank you guys!!! It looks so simple now that you wrote it out.

Subhotosh Khan: That's nice quote, if only if it was true.