and There are 5 white and 3 black balls in one urn, 4 white 8 black balls in second urn and 3 white and 6 black balls in third urn. A ball is taken at random from the first urn and i placed in the second urn, then a ball is taken from the second urn and it is placed in the third urn, and then a ball is taken from the third urn. Find the probability that the third selected ball is white.
Answers
Answer:
The first of three urns contains 5 white and 8 black balls, the second contains white and 4 black balls and the third contains 7 white balls. A persor chooses an urn at random and draws ball from it finds it to be white. The probability that the ball came from second urn is
Step-by-step explanation:
There are three mutually exclusive and exhaustive ways in which 2 balls can be transferred from first bag to second bag.
First way: Two white balls are transferred from first bag to second bag, probability of which is
13
C
2
10
C
2
In the second bag we now have 5 white and 5 black balls and probability of getting a white ball is
10
5
,
∴ Required probability=
13
C
2
10
C
2
×
10
5
=
78
45
×
10
5
=
780
225
Second way: two black balls are transferred from first bag to the second bag
probability of which is
13
C
2
3
C
2
Now we have 3 white and 7 black balls in second bag and probability of getting a white ball is
10
3
∴ Required probability
13
C
2
3
C
2
×
10
3
=
78
3
×
10
3
=
780
9
Third way: One black and one white ball are transferred from first bag to second bag, the probability of which is
13
C
2
10
C
1
×
3
C
1
In second bag, there are now 4 white and 6 black balls and the probability of drawing a white ball is
10
4
∴ Required probability
13
C
2
10
C
1
×
3
C
1
×
10
4
=
78
30
×
10
4
=
780
120
∴ Total Probability =
780
225+9+120
=
780
354