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$


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.


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 $B^c$"

  • $\begingroup$ 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. $\endgroup$ – liya77 May 30 '15 at 14:58
  • $\begingroup$ 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 $\endgroup$ – brumar May 30 '15 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.