Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then, one ball is drawn at random from urn B and placed in urn A. If one ball is drawn at random from urn A, the probability that it is found to be red, is....

Explanation

Fist deraw is from urn A, second draw is from urn B and third draw is again from urn A.
Let veents `A_(R)and A_(B)` be drawing red ball and block ball, respectively, from bag A.
Also, let events `B_(R)and B_(B)` be drawing red ball and black ball, respectively, from bag B.
Case I :
`A_(R)toB_(R)toA_(R)`
Required probability,
`P(A_(R)nnB_(R)nnA_(R))=P(A_(R))P(B_(R)//A_(R))P(A_(R)//(A_(R)nnB_(R)))`
`=6/10xx5/11xx6/10=11/110.`
Case II:
`A_(R)toB_(B)toA_(R)`
Required probability
`P(A_(R)nnB_(B)nnA_(R))=P(A_(R))P(B_(B)//A_(R))P(A_(R)//(A_(R)nnB_(B)))`
`=6/10xx6/11xx5/10=18/110`
Case III :
`A_(B)toB_(R)toA_(R)`
`P(A_(B)nnB_(R)nnA_(R))=P(A_(B))P(B_(R)//A_(B))P(A_(R)//(A_(B)nnB_(R)))`
`=4/10xx4/11xx7/10=56/550`
Case IV :
Required probability
`P(A_(B)nnB_(B)nnA_(R))=P(A_(B))P(B_(B)//A_(B))P(A_(R)//(A_(B)nnB_(B)))`
`=4/10xx7/11xx6/10=84/550`
`therefore` Total probability `=18/110+18/110+56/550+84/550`
`=(90+90+56+84)/(550)=320/550=32/55`