Question

From a box with 5 green, 5 red, 7 black and 3 yellow marbles are drawn 7 marbles at random without replacement. Find the following probabilities:
P(2red,2green,2black,1yellow)
P(the number of black marbles is 5)
When the seven marbles are drawn at random with replacement find
P(2red,2green,2black,1yellow)
P(the number of black marbles is 5)
P(The number of black marbles is amaller than 4)

Answer 1

First find the prob that ALL of them ARE red, which is
(6/11)(5/10)(4/9) = 2/33

So the prob that not all of them are red
= 1-2/33
=31/33

Answer 2

first one is
\[\frac{\dbinom{5}{2}\times \dbinom{5}{2}\times \dbinom{7}{2}\times \dbinom{3}{1}}{\dbinom{20}{7}}\] do you know how to compute these?

Answer 3

yeahh its like

Answer 4

hahaha

Answer 5

second one is
\[\frac{\dbinom{7}{5}\times \dbinom{13}{2}}{\dbinom{20}{7}}\] yes same think just different notation

Answer 6

Thank you!! :)

Answer 7

\[\dbinom{20}{7}=77520\] yw

Answer 8

yeas yeas :)

Answer 9

do you know how to do the ones with replacement? :)