Discussion Week 6 Combinatorics and Conditional Probability

Answer:

• Using the formula for conditional probability for events A and B:

\(P(A | B) = P(A \cap B) / P(B)\) and \(P(B | A) = P(B \cap A) / P(A)\)

we can rewrite the equality in the question to be:

1. \(P(A \cap B)P(A) = P(B \cap A)P(B)\) using cross-multiplication.

• Per the probability communitive law for two events, \(P(A \cap B) = P(B \cap A)\)

thus we can divide the \(P(A \cap B)\) on both sides to get

2. \(P(A) = P(B)\)

• We’re not done yet. We have to show \(P(A) > 1/2\) and to do this, recall the formula

of \(P(A \cup B)\) that is the probability of either of the two events A and B happening

is:

• \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

• We can solve for P(B) to get

• \(P(A \cup B) + P(A \cap B) - P(A) = P(B)\) and plug it into b. to get

• \(P(A) = P(A \cup B) + P(A \cap B) - P(A)\)

• Adding \(P(A)\) on both sides gives

• \(2P(A) = P(A \cup B) + P(A \cap B)\)

• Dividing on both sides by 2 gives

• \(P(A) = P(A \cup B)/2 + P(A \cap B)/2\)

• \(P(A \cup B) = 1\) as per the information given so

• \(P(A) = 1/2 + P(A \cap B)/2\)

• Finally note that the second operand \(P(A \cap B)/2\) will always be within (0,1]

• so \(P(A) \gt 1/2\)