Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 45E from Chapter 2 from Devore's Probability and Statistics for Engineering and Science.
Problem 45E
Chapter:
Problem:
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...
Step-by-Step Solution
Step 1
We are given with probabilities \[\begin{array}{l}P\left( {1 \cap O} \right) = 0.082;\,\,P\left( {1 \cap A} \right) = 0.106;\,\,P\left( {1 \cap B} \right) = 0.008;\,\,P\left( {1 \cap C} \right) = 0.004\,\\P\left( {2 \cap O} \right) = 0.135;\,\,P\left( {2 \cap A} \right) = 0.141;\,\,P\left( {2 \cap B} \right) = 0.018;\,\,P\left( {2 \cap C} \right) = 0.006\,\\P\left( {3 \cap O} \right) = 0.215;\,\,P\left( {3 \cap A} \right) = 0.200;\,\,P\left( {3 \cap B} \right) = 0.065;\,\,P\left( {3 \cap C} \right) = 0.020\end{array}\]
Step 2: (a)
P(A) is sum of all probabilities corresponding to blood group A. That is addition of elements of column 1:\[\begin{array}{l}P\left( A \right) = P\left( {A \cap 1} \right) + P\left( {A \cap 2} \right) + P\left( {A \cap 3} \right)\\ = 0.106 + 0.141 + 0.200\\ = 0.447\end{array}\]Therefore, \[P\left( A \right) = 0.447 \]P(B) is sum of all probabilities corresponding to ethinic group 3. That is addition of elements of row 3:\[\begin{array}{l}P\left( C \right) = P\left( {3 \cap O} \right) + P\left( {3 \cap A} \right) + P\left( {3 \cap B} \right) + P\left( {3 \cap C} \right)\\ = 0.215 + 0.200 + 0.065 + 0.020\\ = 0.50\end{array}\]Therefore, \[P\left( C \right) = 0.50 \]$P\left( {A \cap C} \right)$ is proportion of individuals in ethinic group 3 with blood group A.
Therefore, \[P\left( {3 \cap A} \right) = 0.2\]
Step 3: (b)
$P\left( {A|C} \right)$ is conditional probability of A given C has occurred. \[\begin{array}{l}P\left( {A|C} \right) = \dfrac{{P\left( {A \cap C} \right)}}{{P\left( C \right)}}\\\\ = \dfrac{{0.2}}{{0.5}}\\\\ = 0.4\end{array}\]Therefore, $$P\left( {A|C} \right) = 0.4$$$P\left( {C|A} \right)$ is conditional probability of C given A has occurred. \[\begin{array}{l}P\left( {C|A} \right) = \dfrac{{P\left( {C \cap A} \right)}}{{P\left( A \right)}}\\\\ = \dfrac{{0.2}}{{0.447}}\\\\= 0.447\end{array}\]Therefore, $$P\left( {C|A} \right) = 0.447$$
Step 4: (c)
The probability that selected person does not have type B blood is: \[\begin{array}{l}P\left( {B'} \right) = 1 - P\left( B \right)\\ = 1 - \left( {0.091} \right)\\ = 909\end{array}\]The probability that selected person does is from ethnic group , given that he does not have type B blood is given by:\[\begin{array}{l}P\left( {1|A \cup O \cup AB} \right) = \dfrac{{P\left( {1 \cap \left( {A \cup O \cup AB} \right)} \right)}}{{P\left( {A \cup O \cup AB} \right)}}\\ = \dfrac{{P\left( {\left( {1 \cap A} \right) \cup \left( {1 \cap O} \right) \cup \left( {1 \cap AB} \right)} \right)}}{{P\left( {A \cup O \cup AB} \right)}}\\ = \dfrac{{P\left( {1 \cap A} \right) + P\left( {1 \cap O} \right) + P\left( {1 \cap AB} \right)}}{{P\left( {B'} \right)}}\\ = \dfrac{{0.106 + 0.082 + 0.004}}{{0.909}}\\ = 0.2112\end{array}\]Therefore, \[P\left( {1|A \cup O \cup AB} \right) = 0.2112\]
We are given with probabilities \[\begin{array}{l}P\left( {1 \cap O} \right) = 0.082;\,\,P\left( {1 \cap A} \right) = 0.106;\,\,P\left( {1 \cap B} \right) = 0.008;\,\,P\left( {1 \cap C} \right) = 0.004\,\\P\left( {2 \cap O} \right) = 0.135;\,\,P\left( {2 \cap A} \right) = 0.141;\,\,P\left( {2 \cap B} \right) = 0.018;\,\,P\left( {2 \cap C} \right) = 0.006\,\\P\left( {3 \cap O} \right) = 0.215;\,\,P\left( {3 \cap A} \right) = 0.200;\,\,P\left( {3 \cap B} \right) = 0.065;\,\,P\left( {3 \cap C} \right) = 0.020\end{array}\]
Step 2: (a)
P(A) is sum of all probabilities corresponding to blood group A. That is addition of elements of column 1:\[\begin{array}{l}P\left( A \right) = P\left( {A \cap 1} \right) + P\left( {A \cap 2} \right) + P\left( {A \cap 3} \right)\\ = 0.106 + 0.141 + 0.200\\ = 0.447\end{array}\]Therefore, \[P\left( A \right) = 0.447 \]P(B) is sum of all probabilities corresponding to ethinic group 3. That is addition of elements of row 3:\[\begin{array}{l}P\left( C \right) = P\left( {3 \cap O} \right) + P\left( {3 \cap A} \right) + P\left( {3 \cap B} \right) + P\left( {3 \cap C} \right)\\ = 0.215 + 0.200 + 0.065 + 0.020\\ = 0.50\end{array}\]Therefore, \[P\left( C \right) = 0.50 \]$P\left( {A \cap C} \right)$ is proportion of individuals in ethinic group 3 with blood group A.
Therefore, \[P\left( {3 \cap A} \right) = 0.2\]
Step 3: (b)
$P\left( {A|C} \right)$ is conditional probability of A given C has occurred. \[\begin{array}{l}P\left( {A|C} \right) = \dfrac{{P\left( {A \cap C} \right)}}{{P\left( C \right)}}\\\\ = \dfrac{{0.2}}{{0.5}}\\\\ = 0.4\end{array}\]Therefore, $$P\left( {A|C} \right) = 0.4$$$P\left( {C|A} \right)$ is conditional probability of C given A has occurred. \[\begin{array}{l}P\left( {C|A} \right) = \dfrac{{P\left( {C \cap A} \right)}}{{P\left( A \right)}}\\\\ = \dfrac{{0.2}}{{0.447}}\\\\= 0.447\end{array}\]Therefore, $$P\left( {C|A} \right) = 0.447$$
Step 4: (c)
The probability that selected person does not have type B blood is: \[\begin{array}{l}P\left( {B'} \right) = 1 - P\left( B \right)\\ = 1 - \left( {0.091} \right)\\ = 909\end{array}\]The probability that selected person does is from ethnic group , given that he does not have type B blood is given by:\[\begin{array}{l}P\left( {1|A \cup O \cup AB} \right) = \dfrac{{P\left( {1 \cap \left( {A \cup O \cup AB} \right)} \right)}}{{P\left( {A \cup O \cup AB} \right)}}\\ = \dfrac{{P\left( {\left( {1 \cap A} \right) \cup \left( {1 \cap O} \right) \cup \left( {1 \cap AB} \right)} \right)}}{{P\left( {A \cup O \cup AB} \right)}}\\ = \dfrac{{P\left( {1 \cap A} \right) + P\left( {1 \cap O} \right) + P\left( {1 \cap AB} \right)}}{{P\left( {B'} \right)}}\\ = \dfrac{{0.106 + 0.082 + 0.004}}{{0.909}}\\ = 0.2112\end{array}\]Therefore, \[P\left( {1|A \cup O \cup AB} \right) = 0.2112\]