AP.23 - The Practice of Statistics for AP

Question

AP2.23. There are four major blood types in humans: $\mathrm{O},\mathrm{A},\mathrm{B}$, and $\mathrm{AB}$. In a study conducted using blood specimens from the Blood Bank of Hawaii, individuals were classified according to blood type and ethnic group. The ethnic groups were Hawaiian, Hawaiian-White, HawaiianChinese, and White. Suppose that a blood bank specimen is selected at random.

(a) Find the probability that the specimen contains type $\mathrm{O}$ blood or comes from the Hawaiian-Chinese ethnic group. Show your work.

(b) What is the probability that the specimen contains type AB blood, given that it comes from the Hawaiian ethnic group? Show your work.

(c) Are the events "type B blood" and "Hawaiian ethnic group” independent? Give appropriate statistical evidence to support your answer.

Now suppose that two blood bank specimens are selected at random.

(d) Find the probability that at least one of the specimens contains type A blood from the White ethnic group.

Step-by-Step Solution

Answer

a). The required probability is $0.4516$.

b). The required probability is $0.0212$.

c). They are not independent.

d). The required probability is $0.5706$.

Part (a) Step 1: Given Information

Given data,

Part (a) Step 2: Explanation

To find the probability that the specimen contains type O blood or comes from the Hawaiian-Chinese ethnic group, that

$P(O\cup HC)=P\left(O\right)+P\left(HC\right)-P(O\cap HC)$

$=\frac{62337}{145057}+\frac{5385}{145057}-\frac{2206}{145057}$

$=0.4297+0.03712-0.0152$

$=0.4516$

Part (b) Step 1: Given Information

Given data:

Part (b) Step 2: Explanation

To determine the probability that the samples include type blood AB, assuming that it is from the Hawaiian ethnic group,

$P(AB\mid H)=\frac{P(H\cap AB)}{P\left(H\right)}$

$=\frac{99}{1903+2490+178+99}$

$=0.0212$

Part (c) Step 1: Given Information

Given data,

Part (c) Step 2: Explanation

The probability of having the blood group $B$ is therefore,

$P\left(B\right)=\frac{17604}{145057}$

         $=0.1214$

The probability of Hawaiians having blood group B is as follows:

$P(B\mid \text{ Hawaiian })=\frac{178}{4670}$

                             $=0.0381$

If the two events are independent, their probabilities should be equal. Because the two probability aren't equal, the occurrences "having blood type B" and "Hawaiian ethnic group" aren't mutually exclusive.

Part (d) Step 1: Given Information

Given data:

Part (d) Step 2: Explanation

The chance of not obtaining an A from $W$ is the complement of getting an A from $W$ on the first draw, with a total population of $145057$.

$1-\frac{50008}{145057}$

$=0.6553$

Given that A did not come from W in the initial draw, there are $145057$ persons left, with $50008$ in the A and W group. Because the population was so huge, these two odds are nearly equal.

$1-\frac{50008}{145057}$

$=0.6553$

Use the multiplication rule, as follows, to get the likelihood of both events happening.

$P\left(\text{ None }\right)=0.{6553}^2$

$=0.4294$

We have,

$P(\ge 1)=1-P\left(\text{ None }\right)$

$=1-0.429$

$=0.5706$