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.
​(b) Find the probability of randomly selecting a household that has at least one dog.
(Round to three decimal places as​ needed.) ​
(c) Find the probability of randomly selecting a household that has between one and three​dogs, inclusive.

Answer 1

Answer:

(a) P(x < 2) = 0.871

(b) P(x ≥ 1) = 0.320

(c) P(1 ≤ x ≤ 3) = 0.299

Step-by-step explanation:

We are given the probability distribution of the number of dogs per household in a small town .

Dogs  |   Households  

0        |        0.680

1         |        0.191

2        |        0.079

3        |        0.029

4        |        0.0130

5        |        0.008

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

Fewer than two dogs mean having 0 dog or 1 dog so

P(x < 2) = P(x = 0) + P(x = 1)

P(x < 2) = 0.680 + 0.191

P(x < 2) = 0.871

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

at least one dog means one or greater than one so

P(x ≥ 1) = P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5)

P(x ≥ 1) = 0.191 + 0.079 + 0.029 + 0.0130 + 0.008

P(x ≥ 1) = 0.320

Alternatively:

P(x ≥ 1) = 1 - P(x < 1)

P(x ≥ 1) = 1 - P(x = 0)

P(x ≥ 1) = 1 - 0.680

P(x ≥ 1) = 0.320

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

P(1 ≤ x ≤ 3) = P(x = 1) + P(x = 2) + P(x = 3)

P(1 ≤ x ≤ 3) = 0.191 + 0.079 + 0.029

P(1 ≤ x ≤ 3) = 0.299