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Sagot :
Answer:
1. 0.1684 = 16.84%.
2. 0.2568 = 25.68%
3. 0.0013 = 0.13%
4. 0.0126 = 1.26%.
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they have blood that is Group O, or they do not. The probability of a person having blood that is Group O is independent of any other person, which means that the binomial probability distribution is used 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.
45% of them have blood that is Group O
This means that [tex]p = 0.45[/tex]
Question 1:
This is P(X = 6) when n = 16. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{16,6}.(0.45)^{6}.(0.55)^{10} = 0.1684[/tex]
So 0.1684 = 16.84%.
Question 2:
This is P(X = 3) when n = 8. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{8,3}.(0.45)^{3}.(0.55)^{5} = 0.2568[/tex]
So 0.2568 = 25.68%.
Question 3:
This is P(X = 16) when n = 20. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 16) = C_{20,16}.(0.45)^{16}.(0.55)^{4} = 0.0013[/tex]
So 0.0013 = 0.13%.
Question 4:
This is P(X = 9) when n = 11. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 9) = C_{11,9}.(0.45)^{9}.(0.55)^{2} = 0.0126[/tex]
So 0.0126 = 1.26%.
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