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The probability that a person has a positive rh factor given that he/she has type O blood is approximately 82.2 %
What is the chain rule in probability for two events?
For two events A and B:
The chain rule states that the probability that A and B both occur is given by:
[tex]P(A \cap B) = P(A)P(B|A) = P(B)P( A|B)[/tex]
How to form two-way table?
Suppose two dimensions are there, viz X and Y. Some values of X are there as [tex]X_1, X_2, ... , X_n[/tex] and some values of Y are there as [tex]Y_1, Y_2, ... , Y_n[/tex]
List them in title of the rows and left to the columns. There will be n \times k table of values will be formed(excluding titles and totals), such that:
Value(ith row, jth column) = Frequency for intersection of [tex]X_i[/tex] and [tex]Y_j[/tex] (assuming X values are going in rows, and Y values are listed in columns).
Then totals for rows, columns, and whole table are written on bottom and right margin of the final table.
For n = 2, and k = 2, the table would look like:
[tex]\begin{array}{cccc}&Y_1&Y_2&\rm Total\\X_1&n(X_1 \cap Y_1)&n(X_1\cap Y_2)&n(X_1)\\X_2&n(X_2 \cap Y_1)&n(X_2 \cap Y_2)&n(X_2)\\\rm Total & n(Y_1) & n(Y_2) & S \end{array}[/tex]
where S denotes total of totals, also called total frequency.
n is showing the frequency of the bracketed quantity, and intersection sign in between is showing occurrence of both the categories together.
The missing two-way table for this problem is given below:
[tex]\begin{array}{cccccc}&A&B&AB&O&\rm Total\\\text{Negative}&0.07&0.02&0.01&0.08&0.18\\\text{Positive}{&0.33&0.09&0.03&0.37&0.82\\\rm Total & 0.40&0.11&0.04&0.45&1 \end{array}[/tex]
Instead of frequencies, it contains ratio of frequency to total count of people surveyed (thus, relative frequency).
We want the probability that a person has a positive rh factor given that he/she has type O blood.
If we take:
E = event that a random person chosen has a positive rh factor
and F = event that a random person chosen has type O blood
Then, the needed probability is written symbolically as:
P(E|F).
Using the considered two-way table and the chain rule, we get this probability as:
[tex]P(E|F) = \dfrac{P(E \cap F)}{P(F)} = \dfrac{n(E \cap F)/ n(Total)}{n(F)/n(Total)}[/tex]
Since the two way table consists relative frequency with total count, so, we get:
[tex]n(E \cap F)/n(Total) = 0.37\\n(F)/n(Total) = 0.45[/tex]
[tex]P(E|F) = \dfrac{n(E \cap F)/ n(Total)}{n(F)/n(Total)} = \dfrac{0.37}{0.45} \approx 0.822 = 82.2\%[/tex]
Thus, the probability that a person has a positive rh factor given that he/she has type O blood is approximately 82.2 %
Learn more about two-way table here:
https://brainly.com/question/26788374
Answer:
The answers will be
The probability that a person has a positive Rh factor given that he/she has type O blood is 82 percent.There is a greater probability for a person to have a positive Rh factor given type A blood than a person to have a positive Rh factor given type O blood.
Step-by-step explanation:
Hope this help!
