# [SOLVED] Probability problem check please

• Jan 13th 2009, 09:20 AM
fobos3
Two people are shooting at a target. They both have a chance of 0.2 to hit the target. Find:
(i) What is the probability that they will hit the target at the same time.
(ii) What is the probability that they will hit the target at different times.

I have tried to solve only (i) so far but I suppose the algorithm is the same for (ii). Here is what I have:
After $x+1$ shots for (i) to be true they will have to miss x times and hit it at the x+1 shot.
${(0.8^x \space \times 0.2)}^2 \; x \in [0,\infty]$

So far so good. My problem begins here. They both have a certain chance to hit at $x=n$ but if they don't they have a smaller chance at $x=n+1$ and so on.

So the final probability is: $\sum_{i=0}^{\infty} {(0.8^i \space \times 0.2)}^2$ which is an infinite geometric series. Is that correct?
• Jan 13th 2009, 10:59 AM
Probability
Hello fobos3
Quote:

Originally Posted by fobos3
Two people are shooting at a target. They both have a chance of 0.2 to hit the target. Find:
(i) What is the probability that they will hit the target at the same time.
(ii) What is the probability that they will hit the target at different times.

I assume that the question means that they both shoot together, and continue to shoot at the target until at least one of them hits it. At this point we want to know the probabilities that (i) they both hit at the same time; (ii) one hits and the other misses. Is that how you understand the question?

If that is so, then I would argue as follows:

On any given turn, the probability that they both hit is $0.2 \times 0.2 = 0.04$ and the probability that one hits and one misses is $2 \times 0.2 \times 0.8 = 0.32$. So, on any given occasion, the probability that one hit and one miss occurs is $\frac{0.32}{0.04} = 8$ times greater than the probability that two hits occur.

Now if my assumption about the rules of the game is correct, we can ignore all probabilities until a hit is achieved. But when a hit occurs, the probability that it has been achieved because both have scored a hit is $\frac{1}{9}$ and the probability that one has hit and the other has missed is $\frac{8}{9}$.

• Jan 14th 2009, 12:03 AM
fobos3
Quote:

Hello fobos3I assume that the question means that they both shoot together, and continue to shoot at the target until at least one of them hits it. At this point we want to know the probabilities that (i) they both hit at the same time; (ii) one hits and the other misses. Is that how you understand the question?

If that is so, then I would argue as follows:

On any given turn, the probability that they both hit is $0.2 \times 0.2 = 0.04$ and the probability that one hits and one misses is $2 \times 0.2 \times 0.8 = 0.32$. So, on any given occasion, the probability that one hit and one miss occurs is $\frac{0.32}{0.04} = 8$ times greater than the probability that two hits occur.

Now if my assumption about the rules of the game is correct, we can ignore all probabilities until a hit is achieved. But when a hit occurs, the probability that it has been achieved because both have scored a hit is $\frac{1}{9}$ and the probability that one has hit and the other has missed is $\frac{8}{9}$.

Your assumptions about the game rules are correct. I don't understand a couple of things (sorry really bad at probability).
1. The answers are $\frac{1}{9}$ and $\frac{4}{9}$ for (i) and (ii) respectively.
2. When you get 0.32 for the probability of (ii) why do you multiply by 2? I think it should be 0.16. This affects the ratio $\frac{0.16}{0.04}=4$. Then the probability for (ii) is $\frac{1 \times 4}{9}=\frac{4}{9}$
3. How do you get 1/9 for (i)

Here is how I've solved it.
$\sum_{i=0}^{\infty} {(0.8^i \times 0.2)}^2
=\frac{1}{9}$
see my first post.

Analogically $\sum_{i=0}^{\infty}{(0.8^{i})}^2 \times 0.8 \times 0.2=\frac{4}{9}$

My idea was to I calculate the probability they achieve (i) or (ii) after x misses. Then I had to sum the probabilities for x=0,1,2,...
• Jan 14th 2009, 01:42 AM
mr fantastic
Quote:

Originally Posted by fobos3
Your assumptions about the game rules are correct. I don't understand a couple of things (sorry really bad at probability).
1. The answers are $\frac{1}{9}$ and $\frac{4}{9}$ for (i) and (ii) respectively.
2. When you get 0.32 for the probability of (ii) why do you multiply by 2? I think it should be 0.16. This affects the ratio $\frac{0.16}{0.04}=4$. Then the probability for (ii) is $\frac{1 \times 4}{9}=\frac{4}{9}$
3. How do you get 1/9 for (i)

Here is how I've solved it.
$\sum_{i=0}^{\infty} {(0.8^i \times 0.2)}^2
=\frac{1}{9}$
see my first post.

Analogically $\sum_{i=0}^{\infty}{(0.8^{i})}^2 \times 0.8 \times 0.2=\frac{4}{9}$

My idea was to I calculate the probability they achieve (i) or (ii) after x misses. Then I had to sum the probabilities for x=0,1,2,...

In a given round they either both hit, both miss or one hits and one misses.

Pr(Both hit) = 0.2^2 = 0.04. Pr(Both miss) = 0.8^2 = 0.64.

Therefore Pr(One hits and one misses) = 1 - 0.04 - 0.64 = 0.32 (they are other ways of showing this answer but this is the simplest way to see it I think).

Probability that both hit the target at the same time is:

$0.04 + (0.64)(0.04) + (0.64)^2(0.04) + \, ....$ $= 0.04 (0.64 + 0.64^2 + \, ....) = \frac{4}{100} \cdot \left( \frac{1}{1 - 0.64} \right) = \frac{1}{9}$.

Probability that they hit the target at different times is:

$0.32 + (0.64)(0.32) + (0.64)^2(0.32) + \, ....$ $= 0.32 (0.64 + 0.64^2 + \, ....) = \frac{32}{100} \cdot \left(\frac{1}{1 - 0.64} \right) = \frac{8}{9}$.

This answer makes perfect sense given that the game is played until the target is hit. Obviously the probability that the target is never hit is zero .....

As you can see I completely support the answers found by Grandad. Under our interpretation of the question (the game is played until the target is hit) I don't see how Pr(they hit the target at different times) could possibly be equal to 4/9 since this would mean that the probability that the target is never hit is also 4/9 (clearly false) .....

So either the book is wrong or the question must be interpretted in some other way - a way that's not clear to me.
• Jan 14th 2009, 01:50 AM
Probability
Hello fobos3
Your infinite sum in (i) is correct - although it's not the easiest way - and, with a simple correction, so is your sum in (ii). Read on!
Quote:

Originally Posted by fobos3
1. The answers are $\frac{1}{9}$ and $\frac{4}{9}$ for (i) and (ii) respectively.

I'm a bit puzzled when you say this, if our assumptions about the rules of the game are correct, because, given that a hit has occurred the probabilities for (i) and (ii) must add up to 1. Do you just mean that these are your answers, or have they come from somewhere else?
Quote:

2. When you get 0.32 for the probability of (ii) why do you multiply by 2? I think it should be 0.16. This affects the ratio $\frac{0.16}{0.04}=4$. Then the probability for (ii) is $\frac{1 \times 4}{9}=\frac{4}{9}$
This is the key question. And the answer is simply that either player may hit the target - not just one of them. Look at it like this: call the players A and B. The probability that A hits and B misses is 0.2 x 0.8, and the probability that A misses and B hits is 0.8 x 0.2. These outcomes are independent, so we can add their probabilities together to find the probability that either one or the other occurs. Hence 2 x 0.2 x 0.8. If you use this as part of your infinite sum for part (ii), you also get the answer 8/9.

You might find it helpful to draw a probability tree to illustrate the possible outcomes on one round. See Probabilty Tree Diagrams (with worked solutions & videos) for some examples.
Quote:

3. How do you get 1/9 for (i)
I was trying to avoid something called Bayes' Theorem in answering your question, because many people find the concepts very tricky. If you want to know what it's about, you could try Wikipedia at Bayes' theorem - Wikipedia, the free encyclopedia

So, how did I get 1/9? Assume a hit has occurred. Then it's either because both have hit the target or just one of them has hit and the other missed. There are no other possibilities. As I have said, the probability that on any random round they both hit is 0.04, and the probability that just one hits is 0.32, which is 8 times more likely. So, given that a hit has occurred, we have the following scenario:

P(2 hits) =
$x$
P(1 hit, 1 miss) = $8x$
p(2 hits or 1 hit, 1 miss) = 1, because there are no other possibilities if we know that a hit has occurred.

So
$x + 8x = 1$ and therefore $x = \frac{1}{9}$.

Having done it this way, I think you should be able to see the answer simply by dividing the number 1 in the ratio 1:8.

If you've looked up Bayes' Theorem, then you could say:

• A is the event "Both hit the target on any randomly chosen round". So P(A) = 0.04.
• B is the event "A hit occurs on any randomly chosen round". So P(B) = 0.36 (= 0.04 + 0.32, or, if you prefer, 1 - 0.8x0.8).
• A|B is the event "Both hit the target, given that a hit has occurred". So P(A|B) is what we want to find for (i).
• B|A is the event "A hit occurs, given that both hit the target" Clearly, P(B|A) = 1.

So, using Bayes' Theorem:

$P(A|B)= \frac{P(B|A).P(A)}{P(B)}$

$= \frac{1\times 0.04}{0.36} = \frac{1}{9}$

Are you happy with this now?

• Jan 14th 2009, 05:32 AM
fobos3
Thanks for the reply. Unfortunately there is an error in the problem statement.
(ii) Find the probability that the first person hits the target before the second. Really sorry 'bout that.
• Jan 14th 2009, 06:39 AM
Probability
Hello fobos3
Quote:

Originally Posted by fobos3
Thanks for the reply. Unfortunately there is an error in the problem statement.
(ii) Find the probability that the first person hits the target before the second. Really sorry 'bout that.

In that case, the answer will be $\frac{4}{9}$ - simply because you won't need to multiply by 2 to take the other possibility into account!

• Jan 14th 2009, 03:28 PM
mr fantastic
Quote:

Hello fobos3
In that case, the answer will be $\frac{4}{9}$ - simply because you won't need to multiply by 2 to take the other possibility into account!

It will be 5/9 I think (player going first has to always have the advantage of winning first) ....

Here's an alternative derivation:

Let the probability that first player wins be p. Then the probability that the second player wins is 1 - p. Let the probability of winning a game be r.

First player wins: W + LLW + LLLLW + ....

Second player wins: LW + LLLW + LLLLLW + .... = L(W + LLW + LLLLW + ....)

Therefore $p = (1 - r)p \Rightarrow p = \frac{1}{2 - r}$.

Substitute r = 0.2: p = 5/9.
• Jan 14th 2009, 09:29 PM
Probability
Quote:

Originally Posted by mr fantastic
It will be 5/9 I think (player going first has to always have the advantage of winning first) ....

OK, but this is then impossible...?
Quote:

Originally Posted by fobos3
(i) What is the probability that they will hit the target at the same time.

And...
Quote:

Originally Posted by fobos3
(ii) What is the probability that they will hit the target at different times.

We do seem to be moving the goalposts here. This is surely a new question altogether!

Anyway, I think we've done as much as we can on this one!

• Jan 14th 2009, 09:39 PM
mr fantastic
Quote:

OK, but this is then impossible...?

And...
We do seem to be moving the goalposts here. This is surely a new question altogether!

Anyway, I think we've done as much as we can on this one!

Indeed. I saw the contradiction and drew the same conclusions.
• Jan 15th 2009, 12:26 AM
Probability
At the risk of boring everyone...
Quote:

It will be 5/9 I think (player going first has to always have the advantage of winning first) ....
The problem that Mr F has solved here is this: Two players take turns to aim at a target. The probability that either player hits the target in any given shot is 0.2. The game is over when one player hits the target, and this player wins the game. What is the probability that the player who goes first wins the game?

I thought you might be interested in a recursive solution to this problem, that doesn't involve summing an infinite GP. It's deceptively simple, and goes like this:

Suppose that the probability that the player who goes first wins the game is $x$. Then, since each player is equally likely to hit the target on any given shot, $x$ has the same value whichever player goes first. Suppose now that the players are called A and B, and that A goes first. So the probability that A wins is $x$.

Suppose now that A misses the target at his first attempt. The probability of this event is 0.8. We now effectively have a new game, where B is the player to go first. So the probability that B now wins the game is $x$. Thus the probability that A misses on his first attempt, and that B subsequently wins is $0.8x$.

Now either A wins or B wins. So:

$x + 0.8x = 1$

$\Rightarrow x = \frac{5}{9}$

Neat, eh?