Bag 1 contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball in transferred from Bag 1 to Bag II and them a ball is drawn from Bag II. the ball so drawn is found to be red in colour. Find the probability that the transferred bal is black.

Answer

N/a

Explanation

Let `E_(1)` is the event that a red ball is transferred from bag I to bag II and `E_(2)` is the event that a black ball is transferred from bag I to by II.
Therefore events `E_(1)` and `E_(2)` are mutually exclusive and exhaustive events.
`:.P(E_(1))=3/(3+4)=3/7` and `P(E_(2))=4/(3+4)=4/7`
and let E is the event the ball drawn is red and when a red ball is transferred from bag I to bag II.
Then `P(E/(E_(1)))=(4+1)/((4+1)+5)=5/10=1/2`
when a black ball is transferred from bag I to bag II.
`P(E/(E_(2)))=4/(4+(5+1))=4/10=2/5`
`:.` Required probability
`P((E_(2))/E)=(P(E/(E_(2)))P(E_(2)))/(P(E/(E_(1)))P(E_(1))+P(E/(E_(2)))P(E_(2)))`
`=(2/5xx4/7)/(1/2xx3/7+2/5xx4/7)`
`=(8/35)/(3/14+8/35)=(8/35)/((105+112)/(14xx35))`
`=(8xx14)/217=16/31`