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.
i) Find the chance that the first ball drawn is blue.
ii) Find the chance that the second ball drawn is blue, given that the first ball drawn is blue.
iii) Find the chance that the second ball drawn is blue.
iv) Find the chance that the first ball drawn is blue, given that the second ball drawn is blue.

Answer 1

can you answer (i) ?

Answer 2

This might help. It is very short
http://www.khanacademy.org/math/trigonometry/prob_comb/basic_prob_precalc/v/simple-probability

Answer 3

Yes, (i) is 1/3

Answer 4

ii) Find the chance that the second ball drawn is blue, given that the first ball drawn is blue.

If the first ball that is drawn is blue

then you did this:
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.

so after the first draw, how many blue balls are in the urn? how many balls total are in the urn ?

Answer 5

8 blue ball, total ball is now 18

Answer 6

yes, so problem ii) is done the same way as problem i), except with the new numbers.

Answer 7

thumb up

Answer 8

it is replaced,thus there are 17 balls in the urn and 7 blue balls.
Did I misunderstand?

Answer 9

iii) P(blue given blue) + P(blue given white)= (8/18)*(5/15)+(5/18)*(10/15)
iv) we just go back in time using the forks of Bayes rule: P (first blue given the second blue)= P(2nd blue and 1st blue)/P(2nd blue and 1st blue)+P(2nd blue and 1st white)=
(5/15)*(8/18)/(5/15)*(8/18)+(5/18)*(10/15)=4/9

Answer 10

Thanks for the explanation

Answer 11

That calculation is wrong at dimitrakis

Answer 12

I missed number iv for iii, so sorry dimitrakis

Answer 13

Thanks so much