Let:
A = A student owns a car
B = A student owns a computer
so:
[tex]\begin{gathered} P(A)=0.65 \\ P(B)=0.82 \\ P(A\cap B)=0.55 \end{gathered}[/tex]
a.
A car or computer:
[tex]\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ P(A\cup B)=0.65+0.82-0.55 \\ P(A\cup B)=0.92 \end{gathered}[/tex]
92%
b. A car or a computer but not both:
[tex]\bar{P}(\bar{A}\cap\bar{B})=1-P(A\cap B)=1-0.55=0.45[/tex]
45%
c. A computer given he had a car?
[tex]P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.55}{0.65}\approx0.85[/tex]
85%
d. Had both?
[tex]P(A\cap B)=0.55[/tex]
55%