Question

Use the probability distribution to find probabilities in parts (a) through (c).

The probability distribution of number of dogs per household in a small town

Dogs 0 1 2 3 4 5

Households 0.680 0.191 0.079 0.029 0.0130 0. 008

(a) Find the probability of randomly selecting a household that has fewer than two dogs.

0.871 (Round to three decimal places as needed.)

(b) Find the probability of randomly selecting a household that has at least one dog.

0.320  (Round to three decimal places as needed.)

(c) Find the probability of randomly selecting a household that has between one and three dogs, inclusive.

0.299 (Round to three decimal places as needed.)

Answer 1

(a) Probability of randomly selecting a household that has fewer than two dogs

Probability that household has < 2 dogs = probability of 0 dogs + probability of 1 dog
=  0.680 + 0.191
= 0.871

Probability of randomly selecting a household that has fewer than two dogs = 0.871

(b) Probability of randomly selecting a household that has at least one dog

Probability that household has at least 1 dog = probability that household has > 0 dogs
We know the probability that a household has 0 dogs. So the probability that a household has MORE than 0 dogs means the same as "all possibilities except 0 dogs".

("The probability that x is false = 1 - the probability that x is true". This is called a complement.)
That is = 1 - 0.680
= 0.320

Probability of randomly selecting a household that has at least one dog = 0.320

(c) Probability of randomly selecting a household that has between one and three​ dogs, inclusive

Probability of 1-3 dogs, inclusive = probability of 1 dog + probability of 2 dogs + probability of 3 dogs

= 0.191 + 0.079 + 0.029

= 0.299

   Probability of randomly selecting a household that has between one and three​ dogs, inclusive = 0.299