An urn contains 4 white balls and 6 black balls. A ball is chosen at random, and

Question

An urn contains 4 white balls and 6 black balls. A ball is chosen at random, and its colour noted. The ball is then replaced (placed back to the urn), along with 3 more balls of the same colour (so there are now 13 balls in the urn). Then another bal is drawn at random from the urn 1. Find the chance that the second ball drawn is white. 2. Find the chance that the third ball drawn is black 3. Given that the second ball is drawn white, what is the probability that the first bal drawn s black? 4. Suppose the original content of the urn are w white and b black balls, and that after a ball is drawn from the urn, it is replaced along with d more balls of the same colour. Show that the chance that the second ball is drawn white is 5 Note that w+b the probability agove does not depend on the value of d.

Explanation / Answer

1) probability =P( first ball is white and second ball is white+first ball is black and second ball is white)

=(4/10)*(7/13)+(6/10)*(4/13)=52/130=26/65=0.4

2) probability that third ball is black =P( 1st white*2nd white*3rd black+1st black*2nd white*3rd black+1st white*2nd black*3rd black+1st balck*2nd black*3rd black)

=(4/10)*(7/13)*(6/16)+(6/10)*(4/13)*(9/16)+(4/10)*(10/13)*(13/16)+(6/10)*(9/13)*(12/16)=0.7462

3) probability =P(first ball is black and second ball is white)/P(2nd ball white) =(6/10)*(4/13)/0.4=0.4615

4) probability=P( first ball is white and second ball is white+first ball is black and second ball is white)

=(w/(w+b))*(w+d)/(w+b+d))+(b/(w+b))(w/(w+b+d)) =(1/((w+b)(w+b+d))(w(w+d)+bw)

=(1/((w+b)(w+b+d))(w(w+b+d) =w/(w+b)

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