A fun probability question to try. =]

[brightsky]

2) 48/52*3/51*4/50*3/49*4/48*3/47*40/46*37/45*34/44*31/43*28/42*25/41*22/40 = something pretty small.

Again, I'm doubting my far-fetched logic. :/

[Water]

I'm glad there's not so much combinatorics in year 12 probability!

Me too, my worst area of maths

[kamil9876]

Brightsky, can you please explain how you got the expression for Pr(X=1).

[kevvy]

this question is impossible as probability is NOT fun.

[nacho]

2) 48/52*3/51*4/50*3/49*4/48*3/47*40/46*37/45*34/44*31/43*28/42*25/41*22/40 = something pretty small.

Again, I'm doubting my far-fetched logic. :/

Lol why do you know year 12 probability when you're doing methods next year?
Dr. He disciple i'm assuming..

[brightsky]

Brightsky, can you please explain how you got the expression for Pr(X=1).

Desired outcome/Total number of outcomes

So desired outcome = getting one 4 of a kind and the rest not containing any 4 of a kinds
                           = 11 (since there are 11 possible four of a kinds)*4C4 (one four of a kind) * 48C9 (after getting the four of a kind you only have 48 left out of 52, and from that you choose 9 to make up your 13) --> This is where I am skeptical about the accuracy of my logic, since the subsequent 9 you pick from the remaining 48 can potentially have one or more 4 of a kinds...

*Perhaps* a better shot using the logic I used for question 2 (though doubt it is any more logical than the previous attempt):

44/52*3/51*2/50*1/49*48/48*47/47*46/46*44/45*43/44*42/43*40/42*39/41*38/40 = something ridiculously smaller :/

[kamil9876]

This is where I am skeptical about the accuracy of my logic, since the subsequent 9 you pick from the remaining 48 can potentially have one or more 4 of a kinds...

Yes I was thinking you might've made that mistake. That is why I think that there probably isn't such a simple solution(a product), the simplest would probably be given by the principle of inclusion-exclusion (which is a sum).