Answer:
[tex]\displaystyle \frac{1}{10}[/tex], assuming that the assignment of dormitories is independent from the assignment of dining rooms.
Step-by-step explanation:
Assume that each of the [tex]20\![/tex] dormitories has equal probability of being chosen.
Probability that Kharleen is assigned to one of the [tex]6[/tex] (out of [tex]20[/tex]) dormitories that she likes:
[tex]P(\text{dorm}) = \displaystyle \frac{6}{20}[/tex].
In this fraction, the numerator is the number of dormitories that Khareen likes, while the denominator is the number of all possible dormitories.
Similarly, assume that each of the [tex]6[/tex] dining rooms have equal probability of being chosen.
Probability that Kharleen is assigned to one of the [tex]2[/tex] (out of [tex]6[/tex]) dining rooms that she likes:
[tex]P(\text{dining}) = \displaystyle \frac{2}{6}[/tex].
Similarly, the numerator of this fraction is the number of dining rooms that Khareen likes. The denominator of this fraction is the number of all possible dining rooms.
Assume that the assignment of dormitories and the assignment of dining rooms (two events) are independent.
Hence, the probability of both getting a preferred dormitories (first event) and getting a preferred dining room (second event) would be the product of the probability of the two events:
[tex]\begin{aligned}& P(\text{dorm} \land \text{dining}) \\ &= P(\text{dorm}) \cdot P(\text{dining}) && (\text{by independence}) \\ &= \frac{6}{20} \times \frac{2}{6}= \frac{1}{10}\end{aligned}[/tex].
Hence, the probability that Kharleen is assigned to both a dormitory and a dining room that she likes would be [tex]\displaystyle \frac{1}{10}[/tex] under these assumptions.
