Easy probability questions, plz answer ASAP
 

1) We have to choose a group of 5 people out ot 10(2 are married and 8 others). How many pictures are possible with 5 people if we have to include 1 person out of the married couple in the middle?(and we cant have them both in 1 picture).

2)(Bridge) If we have a hand that consists of 13 cards, how many different hands can we have that contain exactly 3 suits ONLY.

Re: Easy probability questions, plz answer ASAP
 

[ QUOTE ]
1) We have to choose a group of 5 people out ot 10(2 are married and 8 others). How many pictures are possible with 5 people if we have to include 1 person out of the married couple in the middle?(and we cant have them both in 1 picture).


[/ QUOTE ] 2*10!/6! assuming ordering matters

[ QUOTE ]
2)(Bridge) If we have a hand that consists of 13 cards, how many different hands can we have that contain exactly 3 suits ONLY.

[/ QUOTE ] 4*(39!/(26!*13!)) - 3*(26!/(13!*13!)) - 2 assuming you need EXACTLY three, and only the first term if you need three or fewer

This might be wrong. I did it in my head, so who knows?

Re: Easy probability questions, plz answer ASAP
 

[ QUOTE ]
4*(39!/(26!*13!)) - 6*(26!/(13!*13!)) - 4

[/ QUOTE ]Sorry I think this is right on second thought

Re: Easy probability questions, plz answer ASAP
 

2) 4xC(39,13) = 32 489 701 776; therefore, out of a
possible C(52,13) combinations, it is about 5.116379215%.

Re: Easy probability questions, plz answer ASAP
 

[ QUOTE ]
2) 4xC(39,13) = 32 489 701 776; therefore, out of a
possible C(52,13) combinations, it is about 5.116379215%.

[/ QUOTE ]This includes hands like

HHHHHHHHSSSSS

while the question says exactly 3 suits. The 'exactly' qualifier makes it a much more difficult problem than the first one, however, so I'm not sure it is exactly intended.

Re: Easy probability questions, plz answer ASAP
 

Okay, I misread the question. There are:

C(4,2)xC(26,13)= 62 403 600 of AT MOST 2 suits
32 489 701 776 of AT MOST 3 suits

The last number counts EXACTLY 3 suits + AT MOST 2 suits and
the difference is 32 427 298 176.

Re: Easy probability questions, plz answer ASAP
 

[ QUOTE ]
[ QUOTE ]
1) We have to choose a group of 5 people out ot 10(2 are married and 8 others). How many pictures are possible with 5 people if we have to include 1 person out of the married couple in the middle?(and we cant have them both in 1 picture).


[/ QUOTE ] 2*10!/6! assuming ordering matters

[/ QUOTE ]

If order matters it is

2*C(8,4)*4! = 3360

or 2*8!/4! = 3360

That is, 2 ways to pick the married person in the middle, times C(8,4) ways to choose the other 4 people, times 4! ways to order the other 4 people. You used C(10,4). If we only care about who is in the picture and order doesn't matter, then remove the 4! from the first expression to get 140.

Re: Easy probability questions, plz answer ASAP
 

[ QUOTE ]
AT MOST 2 suits

[/ QUOTE ] You're right...subtracting the 4 in my answer double counted the one-suit hands.

[ QUOTE ]
2*8!/4!

[/ QUOTE ] You're also right...I forgot to take away the married people and did 10c4*4!

I suck at life

Re: Easy probability questions, plz answer ASAP
 

yes the 'exactly' was intended [img]/images/graemlins/smile.gif[/img]. Thanks everyone for answering!