# Statistics probability question? (max points)?

You are to pick 4 cards randomly from a condensed deck of cards that contains four suits - ♡,♢,♣,♠ - and the following denomations: Ace, 2, 3, 4, 5, 6, 7, 8, 9, and 10. There are no face-cards in this deck.

What is the probability that you will get

Part (a) a three-of-a-kind?

Part (b) What is the probability that all four cards are of the same suite? For example, all four cards are ♡s?

Part (c) What is the probability you get one-pair?

Part (d) What is the probability you get two aces and two 10s?

Part (e) What is the probability you get two aces and two ♡s?

Please explain how to get any of these answers. Thank you for answering!

### 2 Answers

- PuzzlingLv 76 years agoFavorite Answer
To start with, we will need to know the total number of hands (of any kind) that we can create.

With 40 cards, if we are choosing 4 cards there are "40 choose 4" ways to do that. The "n choose k" formula can be denoted several ways. I'm going to use the notation of C(n,k), where:

C(n,k) = n! / ((n-k)! k!)

C(40,4) = 40 x 39 x 38 x 37 / (4 x 3 x 2 x 1)

= 91390 ways

PART A:

To construct a hand that has 3 of a kind, first pick the rank (A through 10).

C(10,1) = 10 / 1 = 10 ways

From the 4 cards in that rank, pick 3 suits:

C(4,3) = 4 x 3 x 2 / (3 x 2 x 1) = 4 ways

Finally, pick 1 other card from the remaining 36 cards:

C(36,1) = 36 / 1 = 36 ways

Ways to form a 3-of-a-kind:

10 x 4 x 36

= 1440 ways

Answer:

P(three of a kind) = 1440 / 91390

= 144/9139

≈ 1.6%

PART B:

To construct a flush (all the same suit), first pick the suit:

C(4,1) = 4 / 1 = 4 ways

Now from the 10 cards in that suit, pick 4 cards:

C(10,4) = 10 x 9 x 8 x 7 / (4 x 3 x 2 x 1) = 210 ways

Ways to form a hand of all the same suit (flush):

4 x 210

= 840 ways

Answer:

P(flush) = 840 / 91390

= 84/9139

≈ 0.92%

PART C:

To construct a hand with one pair, first pick the rank of the pair (A through 10):

C(10,1) = 10 ways

Pick the two suits for the pair:

C(4,2) = 4 x 3 / (2 x 1) = 6 ways

From the 9 remaining ranks, pick 2 cards that aren't part of a pair:

C(9,2) = 9 x 8 / (2 x 1) = 36 ways

Pick the suit for each card:

C(4,1) x C(4,1) = 16 ways

Ways to form one-pair:

10 x 6 x 36 x 16

= 34560 ways

Answer:

P(one pair) = 34560 / 91390

= 3456/9139

≈ 37.8%

PART D:

Pick the suits for the two aces:

C(4,2) = 4 x 3 / (2 x 1) = 6 ways

Pick the suits for the two tens:

C(4,2) = 4 x 3 / (2 x 1) = 6 ways

Ways to form two aces and two tens:

6 x 6

= 36 ways

Answer:

P(pair of aces, pair of tens) = 36 / 91390

= 18/45695

≈ 0.04%

PART E:

This one isn't fully clear. Are we to assume the aces can't be one of the hearts?

If so, we have *three* suits to pick for the aces.

C(3,2) = 3 x 2 / (2 x 1) = 3 ways

And we have *nine* cards in the hearts (not counting the ace) that we can pick two cards from.

C(9,2) = 9 x 8 / (2 x 1) = 36 ways

Ways to form two aces and two hearts:

3 x 36

= 108 ways

Answer:

P(two aces and two hearts) = 108 / 91390

= 54/45695

≈ 0.12%

- 6 years ago
total 40 cards. picking any card between A and T is 4/40 0.1(10% of chance)

A: divide total amount of possible AAAx(4*3*2*36) combinations by totalcombinations(40x39x38x37) . Around 0.0004 so its around 0.045%

B:Divide amount of all possible XsXsXsXs combinations(10x9x8x7) by total amount of hands. 0.23%

Not gonna count anymore:

C: All possible amount of PPxx combinations(40*3*36*35) by total amount of combos around 7%

D: (4*3*4*3) divide by total combos

E: (3*2*9*8) divide by total combos(if third suited ace doesnt count into the combos)

Im not sure if it works,

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I suppose they want to include the case of the ace of hearts. If so, we can also have 1 ace non-hearts (3 ways), 1 ace of heart (1 way), 1 heart (9 ways) and then a non-ace, non-heart (27 ways). So that would be an additional 729 ways.

(729 + 108) --> 837/91390 ≈ 0.92%