Math, asked by 10premchand98, 1 month ago

There are three bags. The first bag contains 3 white balls, 2 red balls and 4 black
balls, the bag contains 2 white, 3 red and 5 black balls and the third bag contains 3
white, 4 red and 2 black balls. One bag is chosen at random and a ball is drawn. The
drawn ball happens to be a red. What is the probability that the red ball drawn was
taken from (1) first bag, (2) second bag and (3) third bag?

Answers

Answered by Anonymous
0

Step-by-step explanation:

The event of selecting a red ball is denoted by 'R'.

The event of selecting the bag A is denoted by 'A'.

The event of selecting the bag B is denoted by 'B'.

Bag-A has 2 white and 3 red balls

Bag-B has 4 white and 5 red balls

P(A)=P(B)=

2

1

,P(

A

R

)=

5

3

,P(

B

R

)=

9

5

From Bayes' theorem,we get

P(

R

B

)=

P(R)

P(

B

R

)P(B)

=

P(

A

R

)P(A)+P(

B

R

)P(B)

P(

B

R

)P(B)

=

(

5

3

)(

2

1

)+(

9

5

)(

2

1

)

(

9

5

)(

2

1

)

=

52

25

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