Respuesta :
Answer:
a) The probability that fewer than five of them have type o negative blood is 0.1275
b) The probability that fewer than five of them have type o negative blood is 0.9933
c) 0.2708 probability of no donors with type o negative blood. This probability is higher than 0.05, so it would not be unusual having none of the donors with type o negative blood.
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they have type o negative blood, or they do not. The probability of a person having type o negative blood is independent of any other person. 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.
7% of U.S. residents have type o negative blood.
This means that [tex]p = 0.07[/tex]
18 donors.
This means that [tex]n = 18[/tex]
(a) What is the probability that three or more of them have type o negative blood?
Either less than three have, or at least three do. The sum of the probabilities of these events is 1. So
[tex]P(X < 3) + P(X \geq 3) = 1[/tex]
We want [tex]P(X \geq 3)[/tex]
So
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{18,0}.(0.07)^{0}.(0.93)^{18} = 0.2708[/tex]
[tex]P(X = 1) = C_{18,1}.(0.07)^{1}.(0.93)^{17} = 0.3669[/tex]
[tex]P(X = 2) = C_{18,2}.(0.07)^{2}.(0.93)^{16} = 0.2348[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2708 + 0.3669 + 0.2348 = 0.8725[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.8725 = 0.1275[/tex]
The probability that fewer than five of them have type o negative blood is 0.1275
(b) What is the probability that fewer than five of them have type o negative blood?
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{18,0}.(0.07)^{0}.(0.93)^{18} = 0.2708[/tex]
[tex]P(X = 1) = C_{18,1}.(0.07)^{1}.(0.93)^{17} = 0.3669[/tex]
[tex]P(X = 2) = C_{18,2}.(0.07)^{2}.(0.93)^{16} = 0.2348[/tex]
[tex]P(X = 3) = C_{18,3}.(0.07)^{3}.(0.93)^{15} = 0.0942[/tex]
[tex]P(X = 4) = C_{18,4}.(0.07)^{4}.(0.93)^{14} = 0.0266[/tex]
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.2708 + 0.3669 + 0.2348 + 0.0942 + 0.0266 = 0.9933[/tex]
The probability that fewer than five of them have type o negative blood is 0.9933.
c) Would it be unusual f none of the donors had type o negative blood?
[tex]P(X = 0) = C_{18,0}.(0.07)^{0}.(0.93)^{18} = 0.2708[/tex]
0.2708 probability of no donors with type o negative blood. This probability is higher than 0.05, so it would not be unusual having none of the donors with type o negative blood.
