Respuesta :
Answer:
a. 0.042 = 4.2% probability that the chosen applicants are either all managers or all engineers.
b. 0.2398 = 23.98% probability that the number of managers is the same as the number of engineers on the task force.
c. The expected number of engineers chosen is 4.
d. 0.958 = 95.8% probability that at least one manager is chosen for the task force.
Step-by-step explanation:
The positions are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x successes is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this question:
5 + 10 = 15 applicants, which means that [tex]N = 15[/tex]
10 are engineers, which means that [tex]k = 10[/tex]
Six members are chosen, which means that [tex]k = 6[/tex]
a. What is the probability that the chosen applicants are either all managers or all engineers?
Not possible having all managers(five applicants are manager, while there are 6 open positions), so this is P(X = 6), that is, all engineers.
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 6) = h(6,15,6,10) = \frac{C_{10,6}*C_{5,0}}{C_{15,6}} = 0.042[/tex]
0.042 = 4.2% probability that the chosen applicants are either all managers or all engineers.
b. What is the probability that the number of managers is the same as the number of engineers on the task force?
3 engineers, so this is P(X = 3).
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 3) = h(3,15,6,10) = \frac{C_{10,3}*C_{5,3}}{C_{15,6}} = 0.2398[/tex]
0.2398 = 23.98% probability that the number of managers is the same as the number of engineers on the task force.
c. What is the expected number of engineers chosen?
The expected value of the hypergeometric distribution is:
[tex]E(X) = \frac{nk}{N}[/tex]
So
[tex]E(X) = \frac{6(10)}{15} = 4[/tex]
The expected number of engineers chosen is 4.
d. What is the probability that at least one manager is chosen for the task force?
At most five engineers, which is:
[tex]P(X \leq 5) = 1 - P(X = 6)[/tex]
Since in item a. we already have P(X = 6).
[tex]P(X \leq 5) = 1 - 0.042 = 0.958[/tex]
0.958 = 95.8% probability that at least one manager is chosen for the task force.
