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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
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

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yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0well i just got confused.. give me a minute.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay the title of lesson is combinations i got no idea how to do it lol

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0are you sampling with or without replacement?

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0then this is a hypergeometric distribution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a what????? lol explain

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0suppose 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 \[\frac{ _aC_r _bC_g}{_{(a+b)}C_k}\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{_aC_r\cdot_b C_g}{_{a+b}C_k}\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0do you know what \[_nC_r\] is ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope i dont know any of this stuff

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0permutations/combinations?

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0i knwo thats what this unit is called combinations

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0eh right. so it would be like 4/16*7/15 + 7/16*4/15.. right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0id think s if that was the case

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait but youd get the same thing right 7/60 + 7/60

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0then for the second part.. 5/16*4/15.. you get why?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which would equal 14/60 right because you dont change the denominator right or would it be 14/120

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0am i right with the second part @RolyPoly ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well idk how to do this thn i am so lost im gonna fail lol

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0the solution is written above

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay 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
 3 years ago
Best ResponseYou've already chosen the best response.0P(blue first, then green) = P(blue) x P(green after drawing a blue) = 7/16 x 4/(161) = a P(green first, then blue) = P(green) x P(blue after drawing a green) = 4/16 x 7/(161) = b P(1 blue and 1 green) =P(blue first, then green) + P(green first, then blue) = a + b

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now im deffinately confused i just ned a simple explanation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02 red, similar, despite the order of picking a red doesn't matter this time, since you don't the red ones are identical

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what omg i need a tutor

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Which part are you confused at?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02 red actually all of it really

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no thats not the question @phi help please
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