Probability: what are the chances of shooting a target at least once?

[ainster31]

Homework Statement

This is actually from a game that I play online so sorry for the crude question. I have a gun with 3 bullets and 3 targets. The gun randomly shoots the targets. What is the probability that I'll hit target A at least once?

Homework Equations

The Attempt at a Solution

P(hitting target A at least once)
=1 - P(all 3 bullets hitting target A)
=1 - (1/3)*(1/3)*(1/3)
=1 - 1/9
=8/9

So there is a 8/9 probability of hitting target A at least once. I just wanted to make sure that this is correct.

 

[HallsofIvy]

I don't understand your logic. Certainly "all three bullets hitting A" would be an example "hitting A at least once". You should have, rather 1 minus the probability of NO bullets hitting A.
(Also, (1/3)(1/3)(1/3)= 1/27, not 1/9.)

 

[ainster31]

I don't understand your logic. Certainly "all three bullets hitting A" would be an example "hitting A at least once". You should have, rather 1 minus the probability of NO bullets hitting A.
(Also, (1/3)(1/3)(1/3)= 1/27, not 1/9.)

How about now?

P(hitting target A at least once)
=1 - P(no bullets hitting target A)
=1 - (2/3)*(2/3)*(2/3)
=1 - 8/27
=27/27 - 8/27
=19/27

 

[haruspex]

How about now?

P(hitting target A at least once)
=1 - P(no bullets hitting target A)
=1 - (2/3)*(2/3)*(2/3)
=1 - 8/27
=27/27 - 8/27
=19/27

That works.

 

[Mentallic]

Homework Statement

This is actually from a game that I play online so sorry for the crude question. I have a gun with 3 bullets and 3 targets. The gun randomly shoots the targets. What is the probability that I'll hit target A at least once?

This game wouldn't happen to be HEarth'stone? Of the little I've seen of that game, I have witnessed random shots being taken with certain cards being drawn.

You can also extend the problem to shooting k times with n targets. Hitting a specific target at least p times where $0\leq p \leq k$ is described by the binomial distribution function, so depending on whether p is closer to 0 or k depends on which following formula you would use:


$$P(\text{hit specific target at least p times})$$
For p close to 0:
$$= 1-\sum_{i=0}^{p-1}\binom{k}{i}\left(\frac{1}{n}\right)^i\left(1-\frac{1}{n}\right)^{k-i}$$

and for p close to k:
$$= \sum_{i=p}^{k}\binom{k}{i}\left(\frac{1}{n}\right)^i\left(1-\frac{1}{n}\right)^{k-i}$$

 

[mathman523]

If you miss all of them, won't it become 4 possibilities?

 

[ainster31]

@Mentallic: nice, you guessed the right game. Also, it's cool that you generalized it.

 

[Mentallic]

@Mentallic: nice, you guessed the right game. Also, it's cool that you generalized it.

I was a big Blizzard fan back in the Diablo 2 days