Probability of an event is the measure of its chance of occurrence. The correct probabilities are:
Suppose that the sample space is S. Out of which there are two events, A and B, subsets of S.
Then, we can interpret P(A|B) as the probability of that part of A which is in B, with the new sample space as B. (assuming probabililty of B isn't 0).
Symbolically, we write it as:
[tex]P(A|B) = \dfrac{P(A \cap B)}{P(B)}[/tex]
For three sets, A, B and C:
[tex]n(A \cup B \cup C) = n(A) + P(B) + P(C) - n(A\cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)[/tex]
The problem is incomplete. The corresponding venn diagram is attached below.
We can symbolize the venn diagram as:
[tex]n(A) = 12+8+6+5 =31\\n(B) = 11+8+5+3 = 27\\n(C) = 4+8+6+3 = 21\\n(A \cap B) = 5+8=13\\n(B\cap C) = 3+8=11\\n(A \cap C) = 6+8=14\\n(A \cap B \cap C) = 8[/tex]
S = sample space = collection of all unique elements.
S is union of A, B and C.
[tex]n(S) = n(A \cup B\cup C) = 31 + 27 + 21 -13-11-14+8 = 49[/tex]
Checking all the options for their correctness:
We have:
[tex]P(A|C) = P(A\cap C)/P(C) = \dfrac{n(A\cap C)/n(S)}{n(C)/n(S)} = \dfrac{n(A \cap C)}{n(C)} = \dfrac{14}{21} =\dfrac{2}{3}[/tex]
Thus, its correct.
[tex]P(C|B) = P(C\cap B)/P(B) = \dfrac{n(C\cap B)/n(S)}{n(B)/n(S)} = \dfrac{n(C \cap B)}{n(B)} = \dfrac{11}{27} \neq \dfrac{8}{27}[/tex]
Thus, this option is incorrect.
Thus, this option is incorrect.
[tex]P(C) = \dfrac{n(C)}{n(S)}= \dfrac{21}{49} =\dfrac{3}{7}[/tex]
Thus, its correct.
[tex]P(B|A) = P(B\cap A)/P(A) = \dfrac{n(B\cap A)/n(S)}{n(A)/n(S)} = \dfrac{n(B\cap A)}{n(A)} = \dfrac{13}{31} \neq \dfrac{13}{27}[/tex]
Thus, this option is incorrect.
Thus, the correct probabilities are:
Learn more about probability here:
brainly.com/question/1210781
