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

- anonymous

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- anonymous

hey help please?

- yrelhan4

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

- anonymous

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

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## More answers

- anonymous

@ganeshie8

- Zarkon

are you sampling with or without replacement?

- anonymous

without i think

- Zarkon

then this is a hypergeometric distribution

- anonymous

a what????? lol explain

- 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
\[\frac{ _aC_r _bC_g}{_{(a+b)}C_k}\]

- anonymous

yeah i dont get it

- Zarkon

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

- Zarkon

do you know what \[_nC_r\] is ?

- anonymous

not at all

- Zarkon

\[{n\choose r}\]?

- anonymous

nope
i dont know any of this stuff

- Zarkon

permutations/combinations?

- 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?

- anonymous

i knwo thats what this unit is called combinations

- yrelhan4

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

- anonymous

id think s if that was the case

- anonymous

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

- yrelhan4

yup.

- yrelhan4

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

- anonymous

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

- yrelhan4

yeah.

- anonymous

so it is 14/120

- yrelhan4

am i right with the second part @RolyPoly ?

- yrelhan4

it should be 14/120..

- Zarkon

no

- yrelhan4

14/60.. lol

- Zarkon

the answer is 7/30

- anonymous

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

- Zarkon

or 14/60 if you like

- anonymous

how

- Zarkon

the solution is written above

- 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

- 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

- anonymous

now im deffinately confused i just ned a simple explanation

- 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

- anonymous

what omg i need a tutor

- anonymous

Which part are you confused at?

- anonymous

2 BLUE

- anonymous

2 red actually all of it really

- anonymous

u gonna help k?

- anonymous

no thats not the question
@phi help please

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