## Hesan Group Title 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. 2 years ago 2 years ago

1. Hesan Group Title

no

2. Hesan Group Title

two cards, remember

3. Nick2019 Group Title

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. Hesan Group Title

i m getting the wrong answer

5. terenzreignz Group Title

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. Benjammin Group Title

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. Hesan Group Title

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

8. Zarkon Group Title

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

9. Hesan Group Title

10. Hesan Group Title

@Zarkon: yes

11. Zarkon Group Title

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

12. Zarkon Group Title

$\frac{8}{663}$

13. Hesan Group Title

yeah right.. How?

14. Zarkon Group Title

I gave the formula above

15. Zarkon Group Title

this is a classic hypergeometric distribution problem

16. Zarkon Group Title

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. Hesan Group Title

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

18. Zarkon Group Title

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

19. Hesan Group Title

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

20. SmoothMath Group Title

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

21. Zarkon Group Title

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 Group Title

yes...same thing

23. Zarkon Group Title

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

24. SmoothMath Group Title

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. Hesan Group Title

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

26. Hesan Group Title

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

27. Zarkon Group Title

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

28. Hesan Group Title

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 Group Title

that would work...that is what SmoothMath did

30. Hesan Group Title

right. one thing more..

31. Hesan Group Title

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

32. Benjammin Group Title

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

33. Zarkon Group Title

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

34. Hesan Group Title

i m getting it now...

35. Hesan Group Title

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

36. Benjammin Group Title

(4*52)*(3*51)

37. Hesan Group Title

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

38. Zarkon Group Title

yes...if you are drawing without replacement

39. Zarkon Group Title

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

40. Hesan Group Title

alright, thanks.. i get it now!