Question

An urn contains 10 white balls and 5 blue balls. Draws are made repeatedly from the urn as follows. On each draw, a ball is drawn and its color noted; then it is replaced in the urn along with 3 more balls of its color.
1)Find the chance that the second ball drawn is blue.
2)Find the chance that the first ball drawn is blue, given that the second ball drawn is blue.

Answer 1

i think i have the answer:
we taje the two possible assumptions and we apply the additional rule. The probability of the second draw beeing Blue given the first was White is (10+3/18)*(5/18) . The probability of the second draw beeing Blue given the first was Blue is (10/18)*(5+3/18). so the answer is (10+3/18)*(5/18)+(10/18)*(5+3/18)

Answer 2

the answer for the first question that is

Answer 3

Prob. of first ball being white = (10/15) ...when there were total 15 balls
So, the second being blue = (10/15)* (5 / 18) ... when total became 18

Prob. of first ball being blue = (5/15) ...when there were total 15 balls
So, the second being blue = (5/15) * (8/18) ... when total became 18

So, the required probability = (10/15)* (5 / 18) + (5/15) * (8/18)

Answer 4

wil u ans litle bit clear

Answer 5

1)1/3

Answer 6

4c and 4d ?