Probability question worth 10 points?

Question

(c)A bag contains six green, four red and five blue balls. Three balls are selected at random without replacement. Let
C = All three balls are the same color
D = All three balls are green

Find (D|C)

the answer is 10/17 but i do not know which formula i shouls use . please show te working

wyselyn you did not show me the proper way i'm confuse with your answer

Answer 1

Assume, T = total balls

Your numerator for the color x is x!/(x-3)!
so for green it'd be 6!/3! or 6 x 5 x 4

The denominator is T!/(T-3)! = 16 x 15 x 14

But since we're just dealing with D/C, the denominator will
cancels itself out allowing us to just focus on combinations (or the numerator!)

Thus,

D = # of combos of green = 6*5*4

C = # of same color combos = 6*5*4 + 5*4*3 + 4*3*2

So D/C = 6*5*4 / (6*5*4 + 5*4*3 + 4*3*2 )

= 120 / (120+60+24)
= 120 / 204
= 10/17

Answer 2

Yes thats right 10/17 thats what I came up with.
The formula is 12-768/9997tothe 10 over pie

Answer 3

P(D|C) = P(D and C) / P(C)


(6/15)(5/14)(4/13) / ((6/15)(5/14)(4/13) + (4/15)(3/14)(2/13) + (5/15)(4/14)(3/13))

Answer 4

You asked the same question 2 days ago and I answered to that. I obtained the same answer 10/17. You can refer to my working there.

Answer 5

P(D/C)------>This means u should find the conditional probabilty.
P(D/C)=P(D and C occur together)/ P(C)
=P(D n C)/P(C)

Answer 6

d