please do my assignment

a) To get pdf of Y1 or Y2 alone we have to integrate the other variabe completely from joint distribution.
PDF of Y1 =
PDF of Y2 = 6y2-6y22, 0≤y2≤1
b) =P(both)/P(y2<3/4)
See the picture . Our required probability conditional is area of dark red/area of dark red +orange
Area of numerator = area of trapezium = ½ *½ (3/4+1/4) =1/4
Area of denominator = area of triangle with base = ¾ and height =3/4 = 9/32
Hence conditional prob = ¼ *32/9 = 8/9
c) P(Y1/Y2) =...