Suppose that a recent poll of American households about car ownership found that for households with a car, 39% owned a sedan, 33% owned a van, and 7% owned a sports car. Suppose that three households are selected randomly and withreplacement.35)What is the probability that none of the three randomly selected households own a van?

(Round to the nearest thousandth) A)0.060 B)0.699 C)0.036 D)0.301

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JannetPalos JannetPalos · Answer 1 · 2018-10-02T12:16:35+02:00

Let the total number of cars = 100

Total of sedan, van and sports cars = 39+33+7 = 79

Number of other cars = 100 - 79 = 21.

Let A be the event of owning a sedan,

B be the event of owning a van and

C be the event of owning a sports car.

D be the event of owning other cars.

Then, [tex]p(A) = \frac{39}{100}[/tex]

[tex]p(B)=\frac{33}{100}[/tex] and

[tex]p(C) = \frac{7}{100}[/tex]

[tex]p(D) = \frac{21}{100}[/tex]

Now,

p(non selection of van) = p(A ∪ C ∪ D)

= p(A) + p(C) + p(D)

[tex]=\frac{39}{100} +\frac{7}{100} +\frac{21}{100}[/tex]

[tex]=\frac{67}{100}[/tex]

= 0.67

The probability of none of the three selections would be a van is:

0.67 × 0.67 × 0.67

= 0.300763

=0.301

Hence, the correct option is (D).