Chapter 7, Problem 23PT2

(a)

To determine

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

Answer to Problem 23PT2

  $0.4516$ .

Explanation of Solution

In the question it is given there are four major types in humans: O, A, B and AB. In a study conducted using blood specimens for the blood bank of Hawaii are classified according to blood type and ethnic groups such as Hawaiian (H), Hawaiian-White (HW), Hawaiian-Chinese (HC) and white (W). Thus, we have to find the probability that the specimen contains type O blood or comes from the Hawaiian-Chinese ethnic group, that is,

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

Thus, from the table we have that,

	Ethnic group
Blood Type	H	HW	HC	W	Total
O	1903	4469	2206	53759	62337
A	2490	4671	2368	50008	59537
B	178	606	568	16252	17604
AB	99	236	243	5001	5579
Total	4670	9982	5385	125020	145057

Thus, we have,

  $\begin{array}{l}P(O\cup HC)=P(O)+P(HC)-P(O\cap HC)\\ =\frac{62337}{145057}+\frac{5385}{145057}-\frac{2206}{145057}\\ =0.4297+0.03712-0.0152\\ =0.4516\end{array}$

(b)

To determine

To find out what is the probability that the specimen contains type AB blood, given that it comes from the Hawaiian ethic group.

Answer to Problem 23PT2

  $0.0212$ .

Explanation of Solution

In the question it is given there are four major types in humans: O, A, B and AB. In a study conducted using blood specimens for the blood bank of Hawaii are classified according to blood type and ethnic groups such as Hawaiian (H), Hawaiian-White (HW), Hawaiian-Chinese (HC) and white (W). Thus, we have to find the probability that the specimen contains type AB blood, given that it comes from the Hawaiian ethic group, that is,

  $P(AB|H)=\frac{P(H\cap AB)}{P(H)}$

Thus, from the table we have that,

	Ethnic group
Blood Type	H	HW	HC	W	Total
O	1903	4469	2206	53759	62337
A	2490	4671	2368	50008	59537
B	178	606	568	16252	17604
AB	99	236	243	5001	5579
Total	4670	9982	5385	125020	145057

Thus, we have,

  $\begin{array}{l}P(AB|H)=\frac{P(H\cap AB)}{P(H)}\\ =\frac{99}{1903+2490+178+99}\\ =0.0212\end{array}$

(c)

To determine

To find out are the events “type B blood” and “Hawaiian ethnic group” independent.

Answer to Problem 23PT2

They are not independent.

Explanation of Solution

In the question it is given there are four major types in humans: O, A, B and AB. In a study conducted using blood specimens for the blood bank of Hawaii are classified according to blood type and ethnic groups such as Hawaiian (H), Hawaiian-White (HW), Hawaiian-Chinese (HC) and white (W).

Thus, from the table we have that,

	Ethnic group
Blood Type	H	HW	HC	W	Total
O	1903	4469	2206	53759	62337
A	2490	4671	2368	50008	59537
B	178	606	568	16252	17604
AB	99	236	243	5001	5579
Total	4670	9982	5385	125020	145057

Thus, we have,

The probability of having a blood group type B is then,

  $\begin{array}{l}P(B)=\frac{17604}{145057}\\ =0.1214\end{array}$

And the probability of Hawaiians having blood group B is as:

  $\begin{array}{l}P(B|\text{Hawaiian)=}\frac{178}{4670}\\ =0.0381\end{array}$

Thus, the probabilities of both the above two should be identical if the two events are independent. Since the two probabilities are not identical, this then implies that the events “having blood type B” and “Hawaiian ethnic group” are not independent.

(d)

To determine

To find the probability that at least one of the specimens contains type A blood from the White ethnic group.

Answer to Problem 23PT2

  $0.5706$ .

Explanation of Solution

In the question it is given there are four major types in humans: O, A, B and AB. In a study conducted using blood specimens for the blood bank of Hawaii are classified according to blood type and ethnic groups such as Hawaiian (H), Hawaiian-White (HW), Hawaiian-Chinese (HC) and white (W).

Thus, from the table we have that,

	Ethnic group
Blood Type	H	HW	HC	W	Total
O	1903	4469	2206	53759	62337
A	2490	4671	2368	50008	59537
B	178	606	568	16252	17604
AB	99	236	243	5001	5579
Total	4670	9982	5385	125020	145057

Thus, when two specimens are drawn at random, one of 4 things can happen, that is, both are A and W; the first is from A and W, the second is not; The first is not A and W, the second is; Neither are from A and W. These are the only four cases which means their probabilities add up to one. Notice that if we wanted to calculate the probability, case $1-3$ would be all the possible cases and we'd need to calculate them up individually and add together. However, since all probabilities add up to one, doing the trick $1-P\left(\text{case4}\right)$ is a shortcut. Thus, there are total of $145057$ in the population, the probability of not getting an A from W is the complement of getting A from W on the first draw. So,

  $\begin{array}{l}1-\frac{50008}{145057}\\ =0.6553\end{array}$

And given that the first draw did not get A from W, there are $145057$ people remaining and still $50008$ in the A and W group. Note, since the population was large, these two probabilities come out approximately equal to each other. So,

  $\begin{array}{l}1-\frac{50008}{145057}\\ =0.6553\end{array}$

Since we want the probability that both events happen, use the multiplication rule, as follows,

  $\begin{array}{l}\$P\left(\text{None}\right)={0.6553}^2\\ =0.4294\end{array}$

And thus we have,

  $\begin{array}{l}P(\ge 1)=1-P(\text{None)}\\ =1-0.4294\\ =0.5706\end{array}$