Mathematics, 16.04.2020 01:25 alyxkellar06

A study concerning the relative influence of husbands and wives in consumer purchasing, which appeared in recent issue of Time Magazine, reported that husband exerts the primary influence in selecting the make of a new automobile in about 70% of all new-car purchases by families. Suppose five families have decided to buy a new car. a. What is the probability that in exactly two of the five families the husband will exert the primary influence in choosing the make of a car? b. What is the probability that the husband will select the make of car in all five families? c. What is the probability that the husband will exert the primary influence in choosing the make of a car in at least three of the five families? Use Binomial Table given in Appendix B (Text) for this section.

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a) 13.23% probability that in exactly two of the five families the husband will exert the primary influence in choosing the make of a car.

b) 16.81% probability that the husband will select the make of car in all five families

c) 83.69% probability that the husband will exert the primary influence in choosing the make of a car in at least three of the five families

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

Husband exerts the primary influence in selecting the make of a new automobile in about 70% of all new-car purchases by families.

This means that p = 0.7

Suppose five families have decided to buy a new car.

This means that n = 5

a. What is the probability that in exactly two of the five families the husband will exert the primary influence in choosing the make of a car?

This is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.7)^{2}.(0.3)^{3} = 0.1323$

13.23% probability that in exactly two of the five families the husband will exert the primary influence in choosing the make of a car.

b. What is the probability that the husband will select the make of car in all five families?

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.7)^{5}.(0.3)^{0} = 0.1681$

16.81% probability that the husband will select the make of car in all five families

c. What is the probability that the husband will exert the primary influence in choosing the make of a car in at least three of the five families?

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.7)^{3}.(0.3)^{2} = 0.3087$

$P(X = 4) = C_{5,4}.(0.7)^{4}.(0.3)^{1} = 0.3601$

$P(X = 5) = C_{5,5}.(0.7)^{5}.(0.3)^{0} = 0.1681$

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) = 0.3087 + 0.3601 + 0.1681 = 0.8369$

83.69% probability that the husband will exert the primary influence in choosing the make of a car in at least three of the five families

Answer from: Quest

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%

Answer from: Quest

$32,660

step-by-step explanation:

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