Question

you were asked to investigate pets in households (N=44) at a street near the MSU campus. you found 20 houses have cat, 19 houses have dogs, 6 houses have both, and some houses have neither cats or dogs
A. how many classes have neither cats or dogs?
B. if 1 houses is selected at random has dogs, what is the probability that the house also has cats ( statistics or probability problem)?
C.if 1 house is selected at random has cats, what is the probability that the house also has dogs (statistics or probability)?

Answer 1

a) The number of classes have neither cats or dogs

= n(s) - [ n(cat) + n(dogs) - n(both) ]

= 44 - [ 20+19-6] = 11

where n(cat) is number of cats

n(dogs) is number of dogs

b) P( house also has cats / 1 houses is selected at random had dogs)

= P( house also has cats and 1 houses is selected at random had dogs) / P(1 houses is selected at random has dogs)

= (6/44) / (19/44) = 6/19 = 0.3158

c)

P( house also has dogs / 1 houses is selected at random has cats)

= P( house also has dogs and 1 houses is selected at random had cats) / P(1 houses is selected at random has cats)

= (6/44) / (20/44) = 6/20 = 0.3