Urn A contains 1 red ball and 4 blue balls. urn B has 2 red balls and 3 blue balls. in each round you choose at random one ball from urn A and one from urn B. the process is then repeated until the first time you draw two balls of the same color. What is the probability that this process ends on the fourth round?

The probability is 0.06

Step-by-step explanation:

Urn A contains 1 red ball and 4 blue balls.

Urn B contains 2 red balls and 3 blue balls.

The probability of "selecting" a ball of a given colour is equal to the number of balls of that colour in the urn divided by the total number of balls in the urn.

Now, the probability of drawing two balls of the same color in the fourth selection needs to:

1 selection: red ball from A and one blue ball from B, the probability is:

Pa = 1/5, Pb = 3/5, P = 1/5*3/5 = 3/25

second selection: a blue ball from A and one red ball from B, the probability is (remember that now we have one ball less in each urn):

Pa = 4/4, Pb = 2/4, P = 1*2/4 = 1/2

for the third selection we have that we again need to take a blue ball from A and a red ball from B, here the probability is:

Pa = 3/3, Pb = 1/3, P = 1*1/3 = 1/3

the total probability will be the product of the 3 selections

Pt = (1/3)*(1/2)*(3/25) = 0.02

but we have 3 possible permutations of those selections, in one we took the red ball from A in the first selection, other were we took it in the second, and others where we took it in the third.

So the actual probability is 3*Pt = 3*0.02 = 0.06

answer: 298143

step-by-step explanation:

a

step-by-step explanation:

the denominator cannot be the same as one of the factors.

8 denominator times 2 denominator = 16

and 7 numerator time 11 = 77

77/16= 77/16 ( they cannot be further simplified)