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Problem:

In a school competition,the probability of hitting the target bu Dick is $\frac{1}{2}$,by Betty is $\frac{1}{3}$ and by Joe is $\frac{3}{5}$.If all of them fire independently,calculate the probability that the target will be hit.

This general approach for solving is to find the complement of the probability that the target would not be hit.

If the three events are mutually exclusive/independent, then why does summing the three probabilities not give the correct answer?

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    $\begingroup$ The events are independent but not mutually exclusive. $\endgroup$ Dec 7, 2010 at 11:30
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    $\begingroup$ If you sum the probabilities you obtain an answer greater than 1. This provides a quick sanity check that adding them is not the way to go. $\endgroup$ Dec 7, 2010 at 11:38
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    $\begingroup$ "conditional probability" is not the correct phrase here. $\endgroup$
    – Raphael
    Dec 7, 2010 at 11:49
  • $\begingroup$ See my answer in math.stackexchange.com/questions/11415/…. $\endgroup$
    – Shai Covo
    Dec 7, 2010 at 12:22
  • $\begingroup$ More simply, try to understand why for any events $A$ and $B$, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $\endgroup$
    – Shai Covo
    Dec 7, 2010 at 13:13

1 Answer 1

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"Mutually exclusive" and "independent" mean different things.

Two events are mutually exclusive if they can't both happen. For example, "my first name is Steve" and "my first name is Fred" are mutually exclusive. When events are mutually exclusive, you are allowed to add their probabilities to get the probability that one of them occurs.

Independent events are events where finding out about one doesn't change the probability of the other. Finding out that "it is raining" doesn't tell you anything about "my car is red" so those events are independent.

In the question the events "Dick hits the target", "Betty hits the target" and "Joe hits the target" are independent, but are not mutually exclusive. For instance, Betty and Dick could both hit the target. Since they are not mutually exclusive adding the probabilities will not give the correct answer.

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    $\begingroup$ +1 and Accepted,I somehow took the two term in the exact similar sense resulting this confusion.Thank you very much. $\endgroup$
    – Quixotic
    Dec 7, 2010 at 14:57

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