I have been trying to figure out the following question:
A private company has submitted bids on two separate federal government contracts. The company president believes that there is a 45% probability of winning the first contract. If the win the first contract, the probability of the winning the second contract is 70%. However, if they lose the first contract, the president thinks the probability of winning the second contract decreases to 50%. If the above probabilities are true, what is the probability that the company wins only one contract?
This is what I came up with:
Let winning the first contract be event A,
winning the second one be event B,
$P(A)= 0.45, P(A^c)=0.55$
The conditional probability: $P(B|A)=0.70, P(B|A^c)=0.50$
Thus, $P(A\cap$$B)=P(A)\times P(B|A)=0.45 \times 0.70=0.315$
$P(A^c\cap$$B)=P(A^c) \times P(B|A^c)=0.55 \times 0.50=0.275$
$P(B)=P(A\cap$$B) + P(A^c\cap$$B)=0.315+0.275=0.59$
$P(B^c)=1-0.55=0.41$
But now I suppose to find the probability of winning one contract is $P(A^c\cap$$B)$or $P(A \cap$$ B^c)$, how do I find $P(A \cap$$ B^c)$?
I know $P(A \cap$$ B^c)=P(A) \times P(A|B^c)$, then the question becomes how to find the reverse conditional probability of $P(A|B^c)$which is the same as $P(B^c|A)=\frac{P(A \cap B^c)}{P(A)}$.
By the way, the answer to this problem is 0.41.
Update:
brumar has kindly pointed out my mistake: $P(A \cap$$ B^c)=P(A) \times P(A|B^c)$, in fact it should be $P(A \cap$$ B^c)=P(A) \times P(B^c|A)$,
and $P(A|B^c)\neq P(B^c|A)$.