what is the probability that one ends up with exactly 3 jack in a hand of 5 if the first card delt is a jack?

Answer 1

If the first card is Jack, then your remaining cards must be:
2 Jacks out of 3
and 2 others out of 48

No of ways of getting that = 3C2 * 48C2
Total ways of getting the last 4 = 51C4

Ans: 3C2 * 48C2 / 51C4
= 282 / 20825

*EDIT*
No Curt that is not correct. If I assumed the Jacks went into specific positions, I would have got:
3/51 * 2/50 * 48/49 * 47/48 = 47 / 20825
Then I'd have to multiply THAT by 4C2 to get 282/20825.
My answer already incorporates those arrangements.

And you didn't even apply Bayes theorem properly.
A - event that 3Js in first 5 cards
B - event that J is first card
RTF P(A|B) = P(A n B) / P(B)
P(B) = 4/52 = 1/13
P(AnB) = 4/52 * 3/51 * 2/50 * 48/49 * 47/48 * 4C2
You end up with P(A|B) = 282/20825.

Answer 2

There's an error in the first answer above -- implicitly, he assumed that the two jacks went into specific places. So you need to multiply by an additional factor of 4C2=6 to get the correct answer.

Actually, this problem cries out for Bayes' Theorem.

Let A be the event that there are 3 jacks in the first 5 cards
Let B be the event that there is a jack to start with.

P (B contingent A) = 3/5
P (B) = 1/13
P (A) = (4C3 *48C2)/52C5