#1
|
|||
|
|||
Probability involving members in a group
There are six groups of four people each.
There is a deck of cards with aces, twos, threes, fours, fives, and sixes All other cards have been removed. Groups are to be randomly assigned by each individual to pick a card and then depending upon what card they choose are assigned to a group (the aces are a group, twos are a group, ect.) I was wondering the probability of two specific people being in the same group. I think you just set it up like: (4/24)*(3/23)*6 Which is the chance of person A picking a group multiplied by the chance person B picks the same group multiplied by the number of groups. . . I am not all that confident in this answer, and think I am missing something. Thanks for any help. |
#2
|
|||
|
|||
Re: Probability involving members in a group
[ QUOTE ]
There are six groups of four people each. There is a deck of cards with aces, twos, threes, fours, fives, and sixes All other cards have been removed. Groups are to be randomly assigned by each individual to pick a card and then depending upon what card they choose are assigned to a group (the aces are a group, twos are a group, ect.) I was wondering the probability of two specific people being in the same group. I think you just set it up like: (4/24)*(3/23)*6 Which is the chance of person A picking a group multiplied by the chance person B picks the same group multiplied by the number of groups. . . I am not all that confident in this answer, and think I am missing something. Thanks for any help. [/ QUOTE ] Correct. This is just 3/23. The first person can be in any group, and that leaves 3 cards for that group out of 23 total cards, so the probability that the other person is in the same group is 3/23. |
#3
|
|||
|
|||
Re: Probability involving members in a group
Even easier .
Person A already has a number . There are three other players who share the same number and 23 players left . 3/23 |
#4
|
|||
|
|||
Re: Probability involving members in a group
Bruce beat me to it :P
|
#5
|
|||
|
|||
Re: Probability involving members in a group
OP left 28 cards in the deck, not 24 -- but I assume his intention was to remove seven ranks and leave six, in which case 3/23 will indeed be correct.
|
#6
|
|||
|
|||
Re: Probability involving members in a group
[ QUOTE ]
OP left 28 cards in the deck, not 24 -- but I assume his intention was to remove seven ranks and leave six, in which case 3/23 will indeed be correct. [/ QUOTE ] He said "There is a deck of cards with aces, twos, threes, fours, fives, and sixes". That's six ranks. |
#7
|
|||
|
|||
Re: Probability involving members in a group
I read it as there's a deck with those ranks removed....
That is a fine paraphrase of why I only got a 790 on the math SAT [img]/images/graemlins/frown.gif[/img] |
#8
|
|||
|
|||
Re: Probability involving members in a group
Thanks for the help guys. I was pretty sure I was correct, but wanted to check to make sure.
As for the wording, there should have been a period somewhere in there. "There is a deck of cards with aces, twos, threes, fours, fives, and sixes. All other cards have been removed." Maybe that helps a little? |
|
|