## anonymous 3 years ago a jar contains 4 green marbles, 5 red marbles, and 7 blue marbles. If two balls are selected randomly what is the probability of getting : 1. 1 green and 1 blue 2. 2 red 3. 1 blue and 1 red 4. 2 blue

1. anonymous

2. yrelhan4

well i just got confused.. give me a minute.

3. anonymous

okay the title of lesson is combinations i got no idea how to do it lol

4. anonymous

@ganeshie8

5. Zarkon

are you sampling with or without replacement?

6. anonymous

without i think

7. Zarkon

then this is a hypergeometric distribution

8. anonymous

a what????? lol explain

9. Zarkon

suppose you have a box with a red and b green balls. you grab k balls the prob you get r red and g green balls (r+g=k) is given by $\frac{ _aC_r _bC_g}{_{(a+b)}C_k}$

10. anonymous

yeah i dont get it

11. Zarkon

$\frac{_aC_r\cdot_b C_g}{_{a+b}C_k}$

12. Zarkon

do you know what $_nC_r$ is ?

13. anonymous

not at all

14. Zarkon

${n\choose r}$?

15. anonymous

nope i dont know any of this stuff

16. Zarkon

permutations/combinations?

17. yrelhan4

i think it would be easier if we dont use combinations, zarkon.. it would be easier if if we do it like this.. for 1 green ball. 4/16.. 1 blue ball after 1 green ball hass been selected.. 7/15. and then multiply them.. whats your view?

18. anonymous

i knwo thats what this unit is called combinations

19. yrelhan4

eh right. so it would be like 4/16*7/15 + 7/16*4/15.. right?

20. anonymous

id think s if that was the case

21. anonymous

wait but youd get the same thing right 7/60 + 7/60

22. yrelhan4

yup.

23. yrelhan4

then for the second part.. 5/16*4/15.. you get why?

24. anonymous

which would equal 14/60 right because you dont change the denominator right or would it be 14/120

25. yrelhan4

yeah.

26. anonymous

so it is 14/120

27. yrelhan4

am i right with the second part @RolyPoly ?

28. yrelhan4

it should be 14/120..

29. Zarkon

no

30. yrelhan4

14/60.. lol

31. Zarkon

32. anonymous

well idk how to do this thn i am so lost im gonna fail lol

33. Zarkon

or 14/60 if you like

34. anonymous

how

35. Zarkon

the solution is written above

36. anonymous

okay so will you help by showing work on two three and four maybe if i see it on three ill be able to do it

37. anonymous

P(blue first, then green) = P(blue) x P(green after drawing a blue) = 7/16 x 4/(16-1) = a P(green first, then blue) = P(green) x P(blue after drawing a green) = 4/16 x 7/(16-1) = b P(1 blue and 1 green) =P(blue first, then green) + P(green first, then blue) = a + b

38. anonymous

now im deffinately confused i just ned a simple explanation

39. anonymous

2 red, similar, despite the order of picking a red doesn't matter this time, since you don't the red ones are identical

40. anonymous

what omg i need a tutor

41. anonymous

Which part are you confused at?

42. anonymous

2 BLUE

43. anonymous

2 red actually all of it really

44. anonymous

u gonna help k?

45. anonymous

no thats not the question @phi help please