Answer :
let the event of drawing a red ball frm bag 2 be p(A)
p(A)=4/10,
let event of drawing second red ball frm bag 2 is p(B/A)
p(B/A)=3/9
so,the probability of drawing two red balls frm bag 2 is
p(AпB)=4/10.3/9=12/90
(this is the case if the ball transfered frm bag 1 to bag 2 is not red..)
p(A)=4/10,
let event of drawing second red ball frm bag 2 is p(B/A)
p(B/A)=3/9
so,the probability of drawing two red balls frm bag 2 is
p(AпB)=4/10.3/9=12/90
(this is the case if the ball transfered frm bag 1 to bag 2 is not red..)