## anonymous 4 years ago Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

1. anonymous

1/((52)(51)(50))

2. anonymous

1/32600

3. anonymous

4. anonymous

really?

5. anonymous

Yeah ! The answer is 2/5525 But i need to know how these kind of problems are solved

6. anonymous

of well the fact is your chance of the first card being a king is 1/52

7. anonymous

the second card being a king would be 1/51

8. anonymous

and the third card being an ace would be 1/50

9. anonymous

how they came up with that answer i do not know

10. anonymous

unless they are somehow figuring in that there are 4 of each

11. anonymous

Oh Ok there are 4kings So chance of the first card being a king is 4/52

12. anonymous

and then 3/51

13. anonymous

and ace will be 4/50

14. anonymous

So multiply and its 2/5525

15. anonymous

ah you got it...nvm

16. anonymous

But Btw do you know Why we are multiplying ?

17. anonymous

yes

18. anonymous

Um Why ?

19. anonymous

well to find the probability of something happening you have to multiply the number of times it could possibly occure then divide it by the total number of possibilities

20. anonymous

occur

21. anonymous

like if you have 10 books and 2 are school books

22. anonymous

whats the probability that the first book you pull out of your bag would be a school book

23. anonymous

2/10=1/5

24. anonymous

yes But we didnt multiply anything here ?

25. anonymous

so say you have taken 1 school book out already

26. anonymous

whats the chance of pulling both school book out of the bag in a row

27. anonymous

well we already know the first one was 2/10

28. anonymous

well that leaves 1/.9

29. anonymous

so you multiply 2/10 x 1/9

30. anonymous

the probability of pulling them both out in order would be 2/90 or 1/45

31. anonymous

so out of every 90 times you try it, the probability of succedding is 2 times

32. anonymous

OH Ohkay Thanks a lot =)

33. anonymous

we are multiplying because these are independent events. $$P(A\cap B)=P(A) \times P(B)$$

34. anonymous

yes

35. Directrix

Disclaimer: I know this problem has been solved. I am just practicing here. --------------------------------------------------- 4 kings, want 2 4 aces, want 1 no replacement 52 cards in deck P(K) = 4/52. The 52-count drops to 51 with no replacement. King count drops to 3 P(2ndK) = 3/51. The 51 card count drops to 50. King count drops to 2. Ace count remains at 4. P(Ace) = 4 / 50 Using the Multiplication Principle, the probability of these 3 events happening is (4/52) (3/51) (4/ 50) = 2/5525 . Multiplication Principle states: If an event occurs in m ways and another event occurs independently in n ways, then the two events can occur in m × n ways.