In order to calculate the probability of P(Z<3), let's add all cases where Z<3:
[tex]P(Z<3)=P(Z=0)+P(Z=1)+P(Z=2)[/tex]
The minimum value of Z is given when X = 0 and Y = 1, so Z = 1.
The maximum value of Z is given when X = 1 and Y = 2, so Z = 3.
Therefore P(Z = 0) is zero.
Z = 1 can only happen when X = 0 and Y = 1.
Z = 2 can happen when X = 1 and Y = 1 or when X = 0 and Y = 2.
So we can rewrite the expression as follows:
[tex]\begin{gathered} P(Z<3)=0+P(X=0)P(Y=1)+[P(X=1)P(Y=1)+P(X=0)P(Y=2)\rbrack\\ \\ =0+0.5\cdot0.4+0.5\cdot0.4+0.5\cdot0.6\\ \\ =0+0.2+0.2+0.3\\ \\ =0.7 \end{gathered}[/tex]
Therefore the correct option is A.
