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Math Help - A probability question

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    MHF Contributor arbolis's Avatar
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    A probability question

    Imagine an event has a probability of 0.04 to happen. Like for example a gun with a loader with a capacity of 100 bullets but which has only 4 bullets in it. And we push on the thing of the gun to fire a bullet. Say I have 2 identical guns like I just described and I push on the first and then on the second. What is the probability that a bullet was fired? My intuition says a probability close to 0.08 but still less than it and of course greater than 0.04.
    I don't know how I can calculate it. Thanks a lot.
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    Quote Originally Posted by arbolis View Post
    Imagine an event has a probability of 0.04 to happen. Like for example a gun with a loader with a capacity of 100 bullets but which has only 4 bullets in it. And we push on the thing of the gun to fire a bullet. Say I have 2 identical guns like I just described and I push on the first and then on the second. What is the probability that a bullet was fired? My intuition says a probability close to 0.08 but still less than it and of course greater than 0.04.
    I don't know how I can calculate it. Thanks a lot.
    Let A_i be the event that a bullet is fired from the gun 'i', then

    What you want is P(A_1 \cup A_2)...

    Hint1: P(A \cup B) = P(A) + P(B) - P(A \cap B)
    Hint2: A and B are independent events.
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    Quote Originally Posted by arbolis View Post
    Imagine an event has a probability of 0.04 to happen. What is the probability that a bullet was fired?
    It that means at least one bullet is fired the answer is 1 - (.96)^2 =0.078.

    If it means exactly one bullet is fired then 2(0.04)(.96).
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    MHF Contributor arbolis's Avatar
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    A and B are independent, does that mean that P(A \cap B)=0? This is what I recall from what I've learned.
    So P(A \cup B) = P(A) + P(B) ... It seems so strange to me. Does that mean that if I fire exactly 25 guns like I described, then the probability of a bullet to be fired is 1? I don't believe this to be true... I mistaken somewhere.
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Plato View Post
    It that means at least one bullet is fired the answer is 1 - (.96)^2 =0.078.

    If it means exactly one bullet is fired then 2(0.04)(.96).
    Sorry I meant at least 1. Thank you very much. I can now calculate funny math results.
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    Quote Originally Posted by arbolis View Post
    A and B are independent, does that mean that P(A \cap B)=0? This is what I recall from what I've learned.
    So P(A \cup B) = P(A) + P(B) ... It seems so strange to me. Does that mean that if I fire exactly 25 guns like I described, then the probability of a bullet to be fired is 1? I don't believe this to be true... I mistaken somewhere.
    Even though you got your question answered, I'd like to add that P(A \cap B)=0 if they are mutually exclusive, and therefore can not happen at the same time. Not independent.
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    Quote Originally Posted by arbolis View Post
    A and B are independent, does that mean that P(A \cap B)=0? This is what I recall from what I've learned.
    So P(A \cup B) = P(A) + P(B) ... It seems so strange to me. Does that mean that if I fire exactly 25 guns like I described, then the probability of a bullet to be fired is 1? I don't believe this to be true... I mistaken somewhere.
    A and B are independent means P(A \cap B) = P(A) P(B). What you are talking about is disjoint event, also called as mutually exclusive events. They are totally different things.

    Here P(A_1) = P(A_2) = 0.04 and P(A_1 \cap A_2) = P(A_1)P(A_2) = (0.04)^2

    If you substitute the above values, you will get the probability of at least 1 bullet being fired.

    For exactly 1 bullet, compute P((A_1 \cap \overline{A_2}) \cup (A_2 \cap \overline{A_1})), where \overline{A} stands for A complement.
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    Probability

    Hello arbolis
    Quote Originally Posted by arbolis View Post
    Does that mean that if I fire exactly 25 guns like I described, then the probability of a bullet to be fired is 1? I don't believe this to be true... I mistaken somewhere.
    If the probability of any one gun firing is 0.04, and 25 attempts are made, each one independent of the rest, then the probability that none of them fires is (0.96)^{25}.

    So the probability that at least one fires (which I think is what you mean) is 1 - (0.96)^{25} \approx 0.64.

    Grandad

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    MHF Contributor arbolis's Avatar
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    Thank you all!
    If the probability of any one gun firing is 0.04, and 25 attempts are made, each one independent of the rest, then the probability that none of them fires is .

    So the probability that at least one fires (which I think is what you mean) is
    .
    Yes, in fact I asked the question also to solve this! My intuition was thinking it'd be 0.5 and I realized it is close to 0.64... Interesting!
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