a) Do you suppose you can solve the problem if given the suits? Say 4 hearts, 3 clubs, 2 diamonds, and 1 spade?
If so, then just multiply your answer by 4! to allow for the possible arrrangements of the suits.
Ten cards are drawn randomly from a deck of 52 cards (13 cards of each of the 4 different suits).
Each selected card is put into a pile that depends on what suit the card is. (4 total piles)
a) What is the probability that the largest pile contains 4 cards, the next largest pile has 3 cards, the next largest has two cards and the smallest has 1 card?
I am not sure how to begin part a. You want to select 4 of one suit, 3 of a different suit, then 2 of a yet again different suit and then a single card of the last suit. If the numbers were tied to specific suits I could see how to do part a, but this more general case is confusing me.
b) What is the probability that two of the piles have 3 cards, one has 4 cards, and one has no cards?
So when you draw ten cards you don't get 1 of the 4 suits. However, like part a, since the numbers aren't tied to specific suits, I am confused on how to proceed.
Any help would be greatly appreciated. Thank you kindly.
Hello, Sheld!
Here's another approach . . . I'll do part (a).
Ten cards are drawn randomly from a deck of 52 cards.
Each selected card is put into a pile that depends on what suit the card is.
(a) What is the probability that the largest pile contains 4 cards, the next largest
has 3 cards, the next largest has 2 cards and the smallest has 1 card?
There are: . possible 10-card hands.
. . This is the denominator of the probability.
It is not because the order of the cards is not considered.
For the first pile, there are 4 choices for the suit.
. . Then there are ways to get 4 cards of that suit.
For the second pile, there are 3 choices for the suit.
. . Then there are ways to get 3 cards of that suit.
For the third pile, there are 2 choices for the suit.
. . Then there are ways to get 2 cards of that suit.
For the fourth pile, there is 1 choice for the suit.
. . Then there are ways to get 1 card of that suit.
Hence, the numerator is: .