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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.

Answer the following questions:

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 $P(A)+P(B) >1$

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  • $\begingroup$ It seems OK to me. $\endgroup$
    – ajotatxe
    Apr 25, 2015 at 15:01
  • $\begingroup$ "$AB\cap A\cup B$" 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? ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Apr 25, 2015 at 15:11
  • $\begingroup$ @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. $\endgroup$
    – WeakLearner
    Apr 25, 2015 at 15:28
  • $\begingroup$ @MichaelHardy changed it to be consistent though, thanks $\endgroup$
    – WeakLearner
    Apr 25, 2015 at 15:30

1 Answer 1

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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.

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  • $\begingroup$ wouldn't you infer that they are independent from the question ? $\endgroup$
    – WeakLearner
    Apr 25, 2015 at 15:30
  • $\begingroup$ @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${}\,\ldots \qquad{}$ $\endgroup$
    – Michael Hardy
    Apr 25, 2015 at 15:39
  • $\begingroup$ $\ldots\,{}$independent according to a relative-frequency or proportion interpretation of probability. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Apr 25, 2015 at 15:39
  • $\begingroup$ Question regarding this; is there any reason you couldn't use the complement method for this question (for part A)? Would it ever fail? For E.G. 1 - P(ship A missile does NOT hit) * P(ship B missile does NOT hit)? $\endgroup$
    – StormsEdge
    Aug 28, 2018 at 12:51
  • $\begingroup$ @StormsEdge : What you propose is how I would do part (b). But part (b) is not what I was addressing in this answer. $\endgroup$
    – Michael Hardy
    Sep 2, 2018 at 18:07

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