A bag contains 3 red and 7 black balls

Asked by admin @ 17/08/2021 in Math viewed by 297 People

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

Answered by admin @ 17/08/2021

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

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