MAT 230 EXAM TWO
This document is proprietary to Southern New Hampshire University. It and the problems within
may not be posted on any non-SNHU website.
TESTBANKSNERD
1
, Directions: Type your solutions into this document and be sure to show all steps for arriving at
your solution. Just giving a final number may not receive full credit.
PROBLEM 1
This question has 2 parts.
Part 1: Suppose that F and X are events from a common sample space with P (F ) /= 0 and P (X) /= 0.
(a) Prove that P (X) = P (X|F )P (F ) + P (X|F¯)P (F¯). Hint: Explain why P (X|F )P (F ) =
P (X ∩ F ) is another way of writing the definition of conditional probability, and then use
that with the logic from the proof of Theorem 4.1.1.
P (X) = P (X | F )P (F ) + P (X | F )P (F )
As a proof, by definition of conditional probability
P (X ∩ F )
P (X | F ) = P (F ) =⇒ P (X ∩ F ) = P (X | F )P (F )
and
P (X ∩ F )
P (X | F ) = PF ) =⇒ P (∩F ) = P (X | F )P (F )
P (F )
X = (X ∩ F ) ∪ (X ∩ F )
and sets (X ∩ F ) and (X ∩ F ) are disjoint, we have
P (X) = P (X ∩ F ) + P (X ∩ F )
Substituting
P (X) = P (X | F )P (F ) + P (X | F )P (F )
(b) Explain why P (F |X) = P (X|F )P (F )/P (X) is another way of stating Theorem 4.2.1 Bayes
Theorem.
Baye’s theorem states that
P (X | F )P (F )
P (F | X) =
P (X)
By the definition of conditional probability
P (F ∩ X)
P (F | X) =
P (X)
which is Baye’s theorem.
This document is proprietary to Southern New Hampshire University. It and the problems within
may not be posted on any non-SNHU website.
TESTBANKSNERD
1
, Directions: Type your solutions into this document and be sure to show all steps for arriving at
your solution. Just giving a final number may not receive full credit.
PROBLEM 1
This question has 2 parts.
Part 1: Suppose that F and X are events from a common sample space with P (F ) /= 0 and P (X) /= 0.
(a) Prove that P (X) = P (X|F )P (F ) + P (X|F¯)P (F¯). Hint: Explain why P (X|F )P (F ) =
P (X ∩ F ) is another way of writing the definition of conditional probability, and then use
that with the logic from the proof of Theorem 4.1.1.
P (X) = P (X | F )P (F ) + P (X | F )P (F )
As a proof, by definition of conditional probability
P (X ∩ F )
P (X | F ) = P (F ) =⇒ P (X ∩ F ) = P (X | F )P (F )
and
P (X ∩ F )
P (X | F ) = PF ) =⇒ P (∩F ) = P (X | F )P (F )
P (F )
X = (X ∩ F ) ∪ (X ∩ F )
and sets (X ∩ F ) and (X ∩ F ) are disjoint, we have
P (X) = P (X ∩ F ) + P (X ∩ F )
Substituting
P (X) = P (X | F )P (F ) + P (X | F )P (F )
(b) Explain why P (F |X) = P (X|F )P (F )/P (X) is another way of stating Theorem 4.2.1 Bayes
Theorem.
Baye’s theorem states that
P (X | F )P (F )
P (F | X) =
P (X)
By the definition of conditional probability
P (F ∩ X)
P (F | X) =
P (X)
which is Baye’s theorem.
