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

1. 0.1684 = 16.84%.

2. 0.2568 = 25.68%

3. 0.0013 = 0.13%

4. 0.0126 = 1.26%.

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) = (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%.