Question

Assume the box contains 9 balls:
4 red, 2 blue, and 3 yellow. A ball is drawn and its color noted. If the ball is yellow, it is replaced; otherwise, it is not. A second ball is then drawn and its color is noted.
What is the probability that the first ball was yellow, given that the second was red?

Answer 1

8/23

Answer 2

that was right

Answer 3

do you need the step-by-step solution?

Answer 4

Yea really...For now I am trying to just finish what I have left though

Answer 5

the total number of outcomes that have a red ball chosen 2nd are
12 + 20 = 32
the 12, is the total number that have a NON red 1st, with red 2nd
__that comes from 4x3 (4 choices for 1st red x 3 choices for 2nd red)
the 20 is the total number that have red 1st AND 2nd.
__that comes from 5x4 (5 choices for 1st non-red x 4 choices for 2nd red)
of those 32, there are 12 that have yellow 1st with red 2nd.
that comes from 3x4 (3 choices for 1st yellow x 4 choices for 2nd red)
so the probability that the 1st was yellow, GIVEN that the 2nd was red is 12/32

so I don't get 8/23 at all - if you take the trouble to write out the sample space (yes, I did), then I got 12/32

Answer 6

P(Y1 | R2) = P(Y1R2)/P(R2)

P(R2 | Y1) = 4/9 ...............P(Y1) = 1/3
P(R2 | B1) = 1/2 .............P(B1) = 2/9
P(R2 | R1) = 3/8 .............. P(R1) = 4/9

P(R2) = [4/9*1/3] + [1/2*2/9] + [3/8*4/9] = 23/54

P(R2 | Y1) = P(Y1R2)/P(Y1)
P(Y1R2) = P(Y1)*P(R2|Y1) = 1/3* 4/9 = 4/27

--> P(Y1 | R2) = 4/27 * 54/23 = 8/23

Answer 7

Thanks dumb cow