Question

Consider 5 families, each consisting of 4 persons. If it is reported that 6 of the 20 individuals in these families have contagious disease. What is the probability that at lease 3 of the families will be quarantined.

Answer 1

Will it be P(4)+P(5)+P(6) with p = 6/20?

Answer 2

P(at least 3)=1-P(exactly 2 families quarantined) (it is impossible for a single family to be quarantined, since each family only has 4 people)
P(exactly 2 families quarantined) = P(4 from one family, 2 from another)+P(3 from one family, 3 from another)

P(4 from one family, 2 from another) = 5P2*4C4*4C2 / 20C6
P(3 from one family, 3 from another) = 5C2*4C3*4C3 / 20C6

Answer 3

In the above table the eight rows show the possible distributions of the 6 diseased people among the 5 families. There are 5! permutations of the numbers in each row, each permutation having same the number of quarantined families as shown in the right hand column.
There are 6 distributions having 3 or more quarantined families, and 2 distributions having 2 quarantined families. Therefore the probability of 3 or more families being quarantined is:
\[P(3\ or\ more\ families\ quarantined)=\frac{6}{8}=0.75\]