## anonymous 4 years ago Two cards are drawn at random from a deck of well shuffled cards. Determine the probability that both the cards are a king and a queen.

1. anonymous

no

2. anonymous

two cards, remember

3. anonymous

there are 4 suites of 13 cards each , so the probability of getting two cards in each suit is 2/13 there are four suits ,so the probability is 2\13*4 = 8\13 -(use calculator to make it to decimals !!)

4. anonymous

i m getting the wrong answer

5. terenzreignz

I'm not getting the question :( The two cards, are they supposed to be both kings or both queens or one king and one queen?

6. anonymous

Well, there are 52 cards in a deck and 8 of them are Kings and Queens. So the probability of one random draw is 8/52. Assuming you draw one, there are 7/51 remaining, so you just need to multiply that out to get the answer. |dw:1341587716616:dw| ... by my calculations, about 2.11%

7. anonymous

@Benjammin: I did the same thing but getting the wrong answer

8. Zarkon

are you supposed to get one of each card..one king and one queen?

9. anonymous

@Benjammin: your calculations give the answer: 14/663

10. anonymous

@Zarkon: yes

11. Zarkon

$\frac{{4\choose 1}{4\choose1}}{{52\choose 2}}$

12. Zarkon

$\frac{8}{663}$

13. anonymous

yeah right.. How?

14. Zarkon

I gave the formula above

15. Zarkon

this is a classic hypergeometric distribution problem

16. Zarkon

you can look at it this way $$\frac{4}{52}$$ for the first card and $$\frac{4}{51}$$ for the second card but the order we pick them doesn't matter so the answer is $2\cdot\frac{4}{52}\cdot\frac{4}{51}$

17. anonymous

but there shoulb be 8 cards right? 4 for kings and 4 for queens

18. Zarkon

yes..hence the two 4's in the solution

19. anonymous

okey, means 8/52, now how did u get 4/51?

20. anonymous

The way I see it, it's simply: 8/52 * 4/51

21. Zarkon

if you choose the cards from the 8 (4K and 4Q) you might pick up 2 kings or 2 queens, but you only want one of each.

22. Zarkon

yes...same thing

23. Zarkon

pick from the 8...then don't pick a matching card...hence only 4 left to choose from

24. anonymous

Right, for the first card, a king OR a queen is a good outcome. However, assuming you pull one of those, for the second card, only one of those would be a good outcome.

25. anonymous

no i dont get it.. i was thinking it 7/51

26. anonymous

see we've drawn 1 card, now 7 are left, so it would be 7/51

27. Zarkon

but by doing that you could get a matching pair...2 kings or 2 queens

28. anonymous

fine, suppose that on drawing first card, we've got a queen, so we'll exclude queen and now we'll take 4 kings.. am i right?

29. Zarkon

that would work...that is what SmoothMath did

30. anonymous

right. one thing more..

31. anonymous

if we take 4 cards for 2nd turn, will it not be 49 in the denominator? 4/48

32. anonymous

Oooh, I see. What I gave was to draw two cards, each of which would be a king or queen.

33. Zarkon

@Hesan no the experiment is drawing 2 cards from 52....so it is 52 then 51 as above

34. anonymous

i m getting it now...

35. anonymous

okey, now what is the probability that the cards are both jack?

36. anonymous

(4*52)*(3*51)

37. anonymous

the second will always be /51 if we draw 2 cards?

38. Zarkon

yes...if you are drawing without replacement

39. Zarkon

because you pick one of the 52 cards...there are only 51 cards remaining

40. anonymous

alright, thanks.. i get it now!