# Probability of winning one contract

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 $P(A|B^c)\neq P(B^c|A)$.

$P((A\cap B^c) \cup (A^c\cap B))=\\ P(A)P(B^c|A)+P(A^c)P(B|A^c)=\\ 0.45(1-0.7)+(1-0.45)0.5\\=0.41$

To justify the transition from first line to second line remember that

$(A\cap B^c) \cap (A^c\cap B)=\\ (A\cap A^c) \cap (B\cap B^c)=\emptyset$

I know $P(A \cap$$B^c)=P(A) \times P(A|B^c) I think you got it wrong : P(A \cap$$ B^c)=P(A) \times P(B^c|A)$ is the formula you want (and the formula that you need as you know the value of each component).

$P(A|B^c)$which is the same as $P(B^c|A)$

This also incorrect. In a more natural language "now that $B^c$ happened, what is the probability for A" is not the same as "now that $A$ happened, what is the probability for $B^c$"

• So $P(B^c|A)= \frac{P(A \cap B^c)}{P(A)}$ but I don't know what $P(A \cap B^c)$ is. – liya77 May 30 '15 at 14:58
• You should draw a tree to ease your thinking on this kind of problem. Like this amstat.org/publications/jse/v14n1/larsen_figure1.jpg. $P(B^c |A)$ is basically 1-0.45 – brumar May 30 '15 at 15:02