Respuesta :
Answer:
a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message
b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message
c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message
d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message
Step-by-step explanation:
For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.
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}.p^{x}.(1-p)^{n-x}[/tex]
In which [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]
And p is the probability of X happening.
25% of the incoming calls involve fax messages
This means that [tex]p = 0.25[/tex]
25 incoming calls.
This means that [tex]n = 25[/tex]
a. What is the probability that at most 4 of the calls involve a fax message?
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex].
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001[/tex]
[tex]P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006[/tex]
[tex]P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025[/tex]
[tex]P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064[/tex]
[tex]P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118[/tex]
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214[/tex]
0.214 = 21.4% probability that at most 4 of the calls involve a fax message
b. What is the probability that exactly 4 of the calls involve a fax message?
[tex]P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118[/tex]
0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.
c. What is the probability that at least 4 of the calls involve a fax message?
Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So
[tex]P(X < 4) + P(X \geq 4) = 1[/tex]
We want [tex]P(X \geq 4)[/tex]. Then
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001[/tex]
[tex]P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006[/tex]
[tex]P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025[/tex]
[tex]P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064[/tex]
[tex]P(X <4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.001 + 0.006 + 0.025 + 0.064 = 0.096[/tex]
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904[/tex]
0.904 = 90.4% probability that at least 4 of the calls involve a fax message.
d. What is the probability that more than 4 of the calls involve a fax message?
Very similar to c.
[tex]P(X \leq 4) + P(X > 4) = 1[/tex]
From a), [tex]P(X \leq 4) = 0.214)[/tex]
Then
[tex]P(X > 4) = 1 - 0.214 = 0.786[/tex]
0.786 = 78.6% probability that more than 4 of the calls involve a fax message
