Question

The population of a particular country consists of three ethnic groups. Each individual belongs to one of the four major blood groups. The accompanying joint probability table gives the proportions of individuals in the various ethnic group-blood group combinations.

Suppose that an individual is randomly selected from the population, and define events by \({\rm{A = }}\)\{type A selected), \({\rm{B = }}\) (type B selected), and \({\rm{C = }}\) (ethnic group \({\rm{3}}\) selected).

a. Calculate\({\rm{P(A),P(C)}}\), and \({\rm{P(A{C}C)}}\).

b. Calculate both \({\rm{P(A}}\mid {\rm{C)}}\)and \({\rm{P(C}}\mid {\rm{A)}}\), and explain in context what each of these probabilities represents.

c. If the selected individual does not have type B blood, what is the probability that he or she is from ethnic group\({\rm{1}}\)?

Short answer

a) The result is \({\rm{P(A{C}B) = 0}}{\rm{.2}}\).

b) The result is \({\rm{0}}{\rm{.447}}{\rm{.}}\)

c) The probability is \(0.909\)

Step-by-step solution

Definition

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Calculating \({\rm{P(A),P(C)}}\) and \({\rm{P(A{C}C)}}\)

We are given the joint probability table, and we select a random individual. The events are A, B and \({\rm{C}}\)are given in the exercise.

(a):

Probability that types A is selected is the sum of A column of the joint probability table

\({\rm{P(A) = 0}}{\rm{.106 + 0}}{\rm{.141 + 0}}{\rm{.2 = 0}}{\rm{.447}}{\rm{.}}\)

Probability that ethnic group \({\rm{3}}\) is selected is the sum of \({\rm{3}}\) row of the joint probability table

\({\rm{P(C) = 215 + }}{\rm{.2 + }}{\rm{.065 + }}{\rm{.02 = 0}}{\rm{.5}}{\rm{.}}\)

The intersection \({\rm{A{C}B}}\)is the single ethnic group (group \({\rm{3}}\) ) and blood group (type A), which means row \({\rm{3}}\) and column A, or

Therefore, the solution is \({\rm{P(A{C}B) = 0}}{\rm{.2}}\)

Calculating \({\rm{P(A}}\mid {\rm{C)}}\) and \({\rm{P(C}}\mid {\rm{A)}}\)

b)

Conditional probability of A given that the event B has occurred, for which\({\rm{P(B) > 0}}\), is

\({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{P(A{C}B)}}}}{{{\rm{P(B)}}}}\)

for any two event \({\rm{A}}\)and\({\rm{B}}\).

From the definition, we have the following

\({\rm{P(A}}\mid {\rm{C) = }}\frac{{{\rm{P(A{C}C)}}}}{{{\rm{P(C)}}}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.5}}}}{\rm{ = 0}}{\rm{.4}}\)

(1): we have calculated the both probabilities in (a).

This probability represents that if we know that an individual comes from ethnic group \({\rm{3}}\) (event C), the probability that the individual has blood type A (event A) is \({\rm{0}}{\rm{.4}}{\rm{.}}\)

From the definition, we have the following

\({\rm{P(C}}\mid {\rm{A) = }}\frac{{{\rm{P(C{C}A)}}}}{{{\rm{P(A)}}}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{0}}{\rm{.2}}}}{{{\rm{0}}{\rm{.447}}}}{\rm{ = 0}}{\rm{.447}}\)

(1): we have calculated the both probabilities in (a).

This probability represents that if we know that an individual has blood type A (event A), the probability that the individual comes from ethnic group \({\rm{3}}\) (event c) is \({\rm{0}}{\rm{.447}}{\rm{.}}\)

Calculating the probability

c)

The event that individual does not have type B blood is complement of event B defined in the exercise. Let D - "ethnic group \({\rm{1}}\) is selected". We need to find the probability

\(\begin{aligned}P\left( {D\mid {B^\prime }} \right) &= \frac{{P\left( {D \cap {B^\prime }} \right)}}{{P\left( {{B^\prime }} \right)}}\\ &= \frac{{0.192}}{{0.909}}\\ &= 0.211\end{aligned}\)

(1): here we used the fact that

\(P\left( {D \cap {B^\prime }} \right) = 0.082 + 0.106 + 0.004 = 0.192\) which is the probability of being from ethnic group 1 (row l) and not blood type \({\rm{B}}\) (columns \({\rm{O,A}}\) and \({\rm{AB}}\)).

Where for the complement of \({\rm{B}}\)we use the fact that \(0.909\)

\(\begin{aligned}P\left( {{B^\prime }} \right) &= 1 - P(B)\\ &= 1 - (0.008 + 0.018 + 0.065)\\ &= 0.909\end{aligned}\)

where we calculate \({\rm{P(B)}}\) the same way as we did \({\rm{P(A)}}\).