Question

There are 2 urns. 1st urn contains 5 white and 5 blue balls. 2nd urn contains 4 white and 6 black balls. One ball is
taken at random from first urn and put to second urn without noticing its color. Now a ball is chosen at random from
2nd um. What is the probability of the second ball being a bluecolored ball?
choose the best option
10/11
11/13
13/15
none​

Answer 1

Let E

1

​

 be the event of choosing bag I and E

2

​

 be the event of choosing bag II

A be the event of drawing white ball from first bag.

P(E

1

​

)=P(E

2

​

)=

2

1

​

P(A/E

1

​

)=

10

4

​

P(A/E

2

​

)=

9

4

​

By bayes theorem

P(E

1

​

/A)=

P(E

1

​

)P(A/E

1

​

)+P(E

2

​

)P(A/E

2

​

)

P(E

1

​

)P(A/E

1

​

)

​

=

2

1

​

×

10

4

​

+

2

1

​

×

9

4

​

2

1

​

×

10

4

​

​

=

2×9×10

9×4+10×4

​

2

1

​

×

10

4

​

​

=

36+40

4×9

​

=

76

36

​

=

19

9

​

P(E

1

​

/A)=

19

9

​

Answer 2

Answer:

The probability is 6/11.

Step-by-step explanation:

Let P(E) be the probability of second ball being a blue colored ball.

Initially,

• 1st urn: 5 white, 5 blue

• 2nd urn: 4 white, 6 blue

Now, we move one ball from 1st urn to second urn, it has 5/10 possibility of being white and 5/10 possibility of being blue.

• 5/10 (white) - second urn has 6 white, 5 blue in 11 balls
• 5/10 (blue) - second urn has 7 blue, 4 white in 11 balls

P(E) = 5/10 * 5/11 + 5/10 * 7/11

      = 5/22 + 7/22

      = 12/22

      = 6/11.

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