    Question

# Urn A contains six red and four black balls and urn B has four red and six black balls. One ball is drawn at random form urn A and placed in urn B. Then one ball is transfered at random form urn B to urn A. If one ball is now drawn at random from urn A, find the probability that is red. ?

A
3265
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3255
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2355
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5665
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 3255Let Errdenote that a red colour ball is transferred from urn A to urn B tourn then a red colour ball is transferred from urn B to urn A Erb denote that a red colour ball is transferred from urn A to urn B then a black colour ball is transferred from urn B to urn A Ebr denote that a black colour ball is transferred from urn A to urn B then a red colour ball is transferred from urn B to urn A. Ebb denote that a black colour ball is transferred from urn A to urn B then a black colour ball is transferred from urn B to urn A. Then P(Err)=(610)511=311, P(Erb)=(610)611=1855, P(Ebr)=(410)411=85, P(Ebb)=(410)(711)=1455 Let A be the event of drawing a red colour ball after these transfers. Then P(AErr)=610,P(AErb)=510P(AEbr)=710,P(AEbb)=610 Therefore, the requird probability is P(A)=P(Err)P(AErr)+P(Erb)P(AErb) +P(Ebr)P(AEbr)+P(Ebb)P(AEbb)=(311)(610)+(510)1855+(855)(710)+(1455)(610)=90+90+56+84550=3255  Suggest Corrections  0      Similar questions  Related Videos   QUANTITATIVE APTITUDE
Watch in App  Explore more