A probability question

• Jan 15th 2009, 09:07 AM
arbolis
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.
• Jan 15th 2009, 09:13 AM
Isomorphism
Quote:

Originally Posted by arbolis
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 $\displaystyle A_i$ be the event that a bullet is fired from the gun 'i', then

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

Hint1: $\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Hint2: A and B are independent events.
• Jan 15th 2009, 09:24 AM
Plato
Quote:

Originally Posted by arbolis
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 $\displaystyle 1 - (.96)^2 =0.078$.

If it means exactly one bullet is fired then $\displaystyle 2(0.04)(.96)$.
• Jan 15th 2009, 09:25 AM
arbolis
A and B are independent, does that mean that $\displaystyle P(A \cap B)=0$? This is what I recall from what I've learned.
So $\displaystyle P(A \cup B) = P(A) + P(B)$ ... 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.
• Jan 15th 2009, 09:27 AM
arbolis
Quote:

Originally Posted by Plato
It that means at least one bullet is fired the answer is $\displaystyle 1 - (.96)^2 =0.078$.

If it means exactly one bullet is fired then $\displaystyle 2(0.04)(.96)$.

Sorry I meant at least 1. Thank you very much. I can now calculate funny math results.
• Jan 15th 2009, 11:41 AM
Pim
Quote:

Originally Posted by arbolis
A and B are independent, does that mean that $\displaystyle P(A \cap B)=0$? This is what I recall from what I've learned.
So $\displaystyle P(A \cup B) = P(A) + P(B)$ ... 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 $\displaystyle P(A \cap B)=0$ if they are mutually exclusive, and therefore can not happen at the same time. Not independent.
• Jan 15th 2009, 10:41 PM
Isomorphism
Quote:

Originally Posted by arbolis
A and B are independent, does that mean that $\displaystyle P(A \cap B)=0$? This is what I recall from what I've learned.
So $\displaystyle P(A \cup B) = P(A) + P(B)$ ... 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 $\displaystyle P(A \cap B) = P(A) P(B)$. What you are talking about is disjoint event, also called as mutually exclusive events. They are totally different things.

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

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

For exactly 1 bullet, compute $\displaystyle P((A_1 \cap \overline{A_2}) \cup (A_2 \cap \overline{A_1}))$, where $\displaystyle \overline{A}$ stands for A complement.
• Jan 16th 2009, 03:04 AM
Probability
Hello arbolis
Quote:

Originally Posted by arbolis
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 $\displaystyle (0.96)^{25}$.

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