Question

Assume that when blood donors are randomly selected, 45% of them have blood that is Group O (based on data from the Greater New York Blood Program).
1. If the number of blood donors is n = 16 equation, find the probability that the number with Group O blood is equation x = 6.
2. If the number of blood donors is n = 8, find the probability that the number with group O is x = 3.
3. if the number of blood donors is n = 20, find the probability that the number with group O blood is x = 16.
4. if the number of blood donors is n = 11, find the probability that the number with group O blood is x = 9.

Answer 1

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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

45% of them have blood that is Group O

This means that $p = 0.45$

Question 1:

This is P(X = 6) when n = 16. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{16,6}.(0.45)^{6}.(0.55)^{10} = 0.1684$

So 0.1684 = 16.84%.

Question 2:

This is P(X = 3) when n = 8. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{8,3}.(0.45)^{3}.(0.55)^{5} = 0.2568$

So 0.2568 = 25.68%.

Question 3:

This is P(X = 16) when n = 20. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 16) = C_{20,16}.(0.45)^{16}.(0.55)^{4} = 0.0013$

So 0.0013 = 0.13%.

Question 4:

This is P(X = 9) when n = 11. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{11,9}.(0.45)^{9}.(0.55)^{2} = 0.0126$

So 0.0126 = 1.26%.