Answer:
![(2)/(5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/jlb88fidcfo9vyu5ilngpxenga3z24lj93.png)
Explanation:
Given : The proportion of students have a dog :
![P(D)=(45)/(100)](https://img.qammunity.org/2020/formulas/mathematics/high-school/napercmc9ouc8brg2xlc1fw9c42tpur5fb.png)
The proportion of students have a cat :
![P(C)=(30)/(100)](https://img.qammunity.org/2020/formulas/mathematics/high-school/v941frjq7xxj6gys2nvc5d8zkxb8bm5jk0.png)
The proportion of students have both a dog and a cat :
![P(C\cap D)=(18)/(100)](https://img.qammunity.org/2020/formulas/mathematics/high-school/foh25xp625ywn7bxx7wdcypcularmk4zrc.png)
Now, the conditional probability that a student who has a dog also has a cat will be :-
![P(C|D)=(P(C\cap D))/(P(D))\\\\\\\Rightarrow\ P(C|D)=((18)/(100))/((45)/(100))\\\\\\\Rightarrow\ P(C|D)=(18)/(45)=(2)/(5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/5r4ixo4vo68c1c8xsprp43ml3lpsypnjmv.png)
Hence, the probability that a student who has a dog also has a cat =
![(2)/(5)](https://img.qammunity.org/2020/formulas/mathematics/high-school/jlb88fidcfo9vyu5ilngpxenga3z24lj93.png)