## StArAnGeL Group Title 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? 2 years ago 2 years ago

1. ChrisV Group Title

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

2. ChrisV Group Title

1/32600

3. StArAnGeL Group Title

4. ChrisV Group Title

really?

5. StArAnGeL Group Title

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

6. ChrisV Group Title

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

7. ChrisV Group Title

the second card being a king would be 1/51

8. ChrisV Group Title

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

9. ChrisV Group Title

how they came up with that answer i do not know

10. ChrisV Group Title

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

11. StArAnGeL Group Title

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

12. StArAnGeL Group Title

and then 3/51

13. StArAnGeL Group Title

and ace will be 4/50

14. StArAnGeL Group Title

So multiply and its 2/5525

15. gogind Group Title

ah you got it...nvm

16. StArAnGeL Group Title

But Btw do you know Why we are multiplying ?

17. ChrisV Group Title

yes

18. StArAnGeL Group Title

Um Why ?

19. ChrisV Group Title

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. ChrisV Group Title

occur

21. ChrisV Group Title

like if you have 10 books and 2 are school books

22. ChrisV Group Title

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

23. ChrisV Group Title

2/10=1/5

24. StArAnGeL Group Title

yes But we didnt multiply anything here ?

25. ChrisV Group Title

so say you have taken 1 school book out already

26. ChrisV Group Title

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

27. ChrisV Group Title

well we already know the first one was 2/10

28. ChrisV Group Title

well that leaves 1/.9

29. ChrisV Group Title

so you multiply 2/10 x 1/9

30. ChrisV Group Title

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

31. ChrisV Group Title

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

32. StArAnGeL Group Title

OH Ohkay Thanks a lot =)

33. FoolForMath Group Title

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

34. ChrisV Group Title

yes

35. Directrix Group Title

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.