Iron2841

30.10.2021 •

Mathematics

Consider this statement if you are under 17 then you cannot attend this movie

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mobslayer88

12.12.2020

Mathematics

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.

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

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.

7% of U.S. residents have type o negative blood.

This means that p = 0.07

18 donors.

This means that n = 18

(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

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3)$

$P(X \geq 3) = 1 - P(X < 3)$

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

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

$P(X = 0) = C_{18,0}.(0.07)^{0}.(0.93)^{18} = 0.2708$

$P(X = 1) = C_{18,1}.(0.07)^{1}.(0.93)^{17} = 0.3669$

$P(X = 2) = C_{18,2}.(0.07)^{2}.(0.93)^{16} = 0.2348$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2708 + 0.3669 + 0.2348 = 0.8725

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.8725 = 0.1275$

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?

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

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

$P(X = 0) = C_{18,0}.(0.07)^{0}.(0.93)^{18} = 0.2708$

$P(X = 1) = C_{18,1}.(0.07)^{1}.(0.93)^{17} = 0.3669$

$P(X = 2) = C_{18,2}.(0.07)^{2}.(0.93)^{16} = 0.2348$

$P(X = 3) = C_{18,3}.(0.07)^{3}.(0.93)^{15} = 0.0942$

$P(X = 4) = C_{18,4}.(0.07)^{4}.(0.93)^{14} = 0.0266$

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

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?

$P(X = 0) = C_{18,0}.(0.07)^{0}.(0.93)^{18} = 0.2708$

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.

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