A study concerning the relative influence of husbands

Ответ:

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24.01.2022

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

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