Question

A bag contains 3 red and 7 black balls. Two balls are selected at random with replacement. If second selected is given to be red, what is the probability that first selected is also red

Answer 1

Answer:

2/9

Step-by-step explanation:

Let:

• R1 be the event that the first ball is red
• B1 be the event that the first ball is black
• R2 be the event that the second ball is red

We are asked for the conditional probability P( R1 | R2 ).

P( R1 | R2 ) = P( R1 ∩ R2 ) / P( R2 )

                = P( R2 | R1 ) P( R1 ) / P( R2 )             ... (*)

Deal with the numerator and denominator here separately.

Numerator

P( R1 ) = ( # red balls at start ) / ( # balls at start )

         = 3 / 10

P( R2 | R1 ) = ( # red balls after selecting one ) / ( # balls after selecting one )

                = 2 / 9

So the numerator is:

3/10  ×  2/9  =  1 / 15

Denominator

By the Law of Total Probability,

P( R2 ) = P( R2 | R1 ) P ( R1 )  +  P( R2 | B1 ) P( B1 )

The first term here is just the numerator that we worked out above.

For the second term:

P( B1 ) = ( # balck balls at start ) / ( # balls at start )

         = 7 / 10

P( R2 | B1 ) = ( # red balls after selecting black ) / ( # balls after selecting one )

                = 3 / 9   =  1 / 3

Thus...

P ( R2 ) = 1/15  +  7/10 × 1/3

           = 2/30 + 7/30

           = 9 / 30

           = 3 / 10

Putting it together

Going back to equation (*), we now have

P( R1 | R2 ) = P( R2 | R1 ) P( R1 ) / P( R2 )

                = ( 1/15 ) / ( 3/10 )

                = 10 / ( 15 × 3 )

                = 10 / 45

                = 2 / 9

Answer 2

Hope this will helps you