# Conditional Probability of Sinking Ship Question

Question: Two ships. Ship A's missiles have an 80% probability of hitting its target, ship B's missiles have a 50% probability of hitting the target. It only takes one hit from a missile to sink a submarine.

a) Both ships are aiming at the same submarine, and both fire a missile. What is the probability that the submarine sinks.

b) Given that the submarine is seen sinking, what is the probability that both missiles hit.

Attempt:

(a) $P(A \cup B) = P(A) + P(B) - P(A\cap B) = P(A) + P(B) - P(A) P(B) = 1.3 - 0.4 = 0.9$

(b) $P ( A \cap B \mid A \cup B) = \dfrac{P(A\cap B \cap (A\cup B) )}{P(A \cup B)}=\dfrac{P(A \cap B)}{P(A)+P(B)-P(A)P(B)} = \dfrac{4}{9}$

Are these correct? Im confused how to treat it when $P(A)+P(B) >1$

• It seems OK to me. Apr 25, 2015 at 15:01
• "$AB\cap A\cup B$" is ambiguous at best. Is it $\Big(AB\cap A\Big) \cup B$ or is it $AB\cap\Big(A\cup B\Big)$? And why use two different notations for intersection, one with "$\cap$" on one expressed by juxtaposition? ${}\qquad{}$ Apr 25, 2015 at 15:11
• @MichaelHardy I guess I use the juxtaposition usually, but when it comes to having an expression that already has a cap in it i wanted to make it most clear. Apr 25, 2015 at 15:28
• @MichaelHardy changed it to be consistent though, thanks Apr 25, 2015 at 15:30

It is correct if $A$ and $B$ are independent. The fact that $P(A)+P(B)>1$ doesn't upset anything. But see my comments on notation above.
• @dimebucker91 : That's a somewhat subtler question than it appears. Probably I would treat them that way, although I prefer to be explicit about that. If the probability of having blue eyes must be construed as the proportion of the population that has blue eyes, then one cannot say the probability that there was life on Mars a billion years ago is $1/2$, since we cannot say it was true in half of all cases. BUT if we allow such purely epistemic probabilities as the latter, then two propositions might be conditionally independent given what is known even if they're not${}\,\ldots \qquad{}$ Apr 25, 2015 at 15:39
• $\ldots\,{}$independent according to a relative-frequency or proportion interpretation of probability. ${}\qquad{}$ Apr 25, 2015 at 15:39