Blood types statistics

Problem

The following table depicts the distribution of blood types and rhesus factors amongst the german population:

	0	A	B	AB
Rh+	35%	37%	9%	4%
Rh-	6%	6%	2%	1%

During a morning, 25 people donate blood in a hospital. In the following it shall be assumed that those 25 people represent a random sample of the population.

1. Determine the probability that exactly ten of the donors have blood type A.

2. Find the probability that more than half of the donors have blood type A and rhesus factor Rh+.

The following table shows which donor blood is suitable for the different recipients:

		Donor
		0 Rh-	0 Rh+	A Rh-	A Rh+	B Rh-	B Rh+	AB Rh-	AB Rh+
Recipient	AB Rh+	✓	✓	✓	✓	✓	✓	✓	✓
	AB Rh-	✓		✓		✓		✓
	B Rh+	✓	✓			✓	✓
	B Rh-	✓				✓
	A Rh+	✓	✓	✓	✓
	A Rh-	✓		✓
	0 Rh+	✓	✓
	0 Rh-	✓

3. A patient with blood type B and rhesus factor Rh- needs donor blood. Determine how many randomly selected people have to donate blood in order to obtain at least one suitable donor blood with a probability higher than 95 %.

Solution of part a

The probability to find exactly ten people with blood type A from 25 blood donors is given by the binomial distribution. The probability for each single donor to have blood type A is:

\[P(\mathrm{A}) = P(\mathrm{A, Rh+}) + P(\mathrm{A, Rh-}) = 43 \%\,.\]

The desired probability is:

\[\begin{split}W^{25}_{0.43}(10) =& {25 \choose 10}\cdot0{.}43^{10}\cdot(1-0{.}43)^{15}\\ =& \frac{25!}{10!\cdot 15!} 0{.}43^{10} \cdot 0{.}57^{15} \approx 15{.}4 \% \,.\end{split}\]

This random experiment can be simulated with Sage:

Solution of part b

The probability that more than half of the donors have blood type 0 and rhesus factor Rh+ can easily be determined with Sage by summation:

Solution of part c

According to the table, people with blood type 0 Rh- as well as B Rh- can donate to a recipient with blood type B and rhesus factor Rh-. The probability to find a suitable donor thus equals

\[P(\mathrm{0, Rh-}) + P(\mathrm{B, Rh-}) = 8\%\,.\]

The probability that a person is not a suitable donor then equals \(92\%\). The probability that there is no suitable donor amongst \(n\) people thus amounts to \(0.92^n\). We look for the smallest number \(n\) for which

\[(0{.}92)^{n} \leq 0{.}05\,.\]

Taking the logarithm and bearing in mind that \(\ln(0.92)\) is negative, one finds

\[n \geq \frac{\ln(0{.}05)}{\ln(0{.}92)} \approx 35{.}9\,.\]

One hence needs at least 36 donors.

With the help of a random experiment one can approximately determine with Sage the probability that amongst 36 donors there is at least one suitable donor.