Answer :
Answer:
There is a 5.56% probability that there is at one man in this sample
Step-by-step explanation:
For each nanny, there are only two possible outcomes. Either it is a women, or it is a men. This means that we can solve this problem using the binomial probability distribution.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And [tex]\pi[/tex] is the probability of X happening.
Of 4,176 nannies, only 24 were men. This means that [tex]\pi = \frac{24}{4176} = 0.0057[/tex].
Now find the probability that in a random sample of 10 nannies who were placed last year, at least 1 is a man
Sample of 10 nannies, so [tex]n = 10[/tex]
Either there is at least one nanny that is a men, that is probability [tex]P(X > 0)[/tex], or there are no nannies there are men, that is probability [tex]P(X = 0)[/tex]. The sum of these probabilities is decimal 1. We want to find [tex]P(X>0)[/tex].
[tex]P(X > 0) + P(X = 0) = 1[/tex]
[tex]P(X > 0) = 1 - P(X = 0)[/tex]
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.0057)^{0}.(0.9943)^{10} = 0.9444[/tex]
[tex]P(X > 0) = 1 - P(X = 0) = 1 - 0.9444 = 0.556[/tex]
There is a 5.56% probability that there is at one man in this sample