# Probability: B needs to shoot more times than A

1. Jun 17, 2010

### SamTaylor

Hi

I am having some problems solving this exercise. Can somebody give a hint on how to solve this. The hint from the book is not really helping me.

1. The problem statement, all variables and given/known data

Two sharpshooters, A and B, are going to shoot at a target. A has probability Pa of hitting it on a single shot; B has probability Pb of hitting it on a single shot. Wheater the target is struck on any one shot is statistically independent of whether it is struck on any other shot. What is the probability that B needs to shoot more times before hitting the target than A?

(Hint from the book:
1. Suppose that A hits the target for the first time on his nth shot.
2. Calculate the probabity that B shoots more than n times before hitting the target
3. Then use the principle of total probabity to account for all values of n from 1 ad infinium)

2. Relevant equations

Pr(M AND M) = Pr(M|Ai)*Pr(Ai)
Pr(M) = Pr(M|A1)*Pr(A1) + Pr(M|A2)*Pr(A2) ...

3. The attempt at a solution

P(A nth shot) = Pa*(1-Pa)^(n-1)
P(B nth+1 shot) = Pb*(1-Pb)^n
????

Thank you

2. Jun 17, 2010

### vela

Staff Emeritus
You could think of it this way. In each trial, each shooter keeps shooting until he hits the target. Say A hits the target on shot nA and B hits the target on shot nB. You have a bunch of possible outcomes (nA,nB). The outcome (1,1) would mean they both hit on their first shot; (1,2) would mean A hits on the first shot and B hits on his second shot; (2,1) would mean A needed two shots and B only one; and so on. What's the probability P(nA,nB) of outcome (nA,nB)?

Finally, the event you're interested in consists of the outcomes where nB>nA. How would you express that in terms of the probabilities of the individual outcomes?

3. Jun 20, 2010

### SamTaylor

> What's the probability $$P(n_A,n_B)$$ of outcome $$(n_A,n_B)$$?

$$P(n_a,n_b) = P_a*(1-P_a)^{(n_a-1)}*P_b*(1-P_b)^{(n_b-1)}$$

> Finally, the event you're interested in consists of the outcomes where $$n_B>n_A$$.
> How would you express that in terms of the probabilities of the individual outcomes?

I have no clue.

4. Jun 20, 2010

### vela

Staff Emeritus
Then you better go back and review basic probability.

5. Jun 20, 2010

### SamTaylor

> Then you better go back and review basic probability.

I already did. I guess here is some point i don't understand.
So is the first assumption i did for $$P(n_a,n_b)$$ also wrong?

6. Jun 20, 2010

### vela

Staff Emeritus
Your expression for P(nA,nB) is correct.

7. Jun 23, 2010

### SamTaylor

with the help of
$$\sum_{n=0}^{\infty} q^n = \frac{1}{1-q}$$

$$P(b > a) = \sum_{n_a}^{\infty} \sum_{n_b=n_a+1}^{\infty} P(a_n,a_b) = 1-P_b$$

is this correct ?

8. Jun 23, 2010

### vela

Staff Emeritus
No, it's not correct. Your answer should depend on pA somehow. Let pB be fixed. If pA>pB, you'd expect A would typically hit the target before B does. If pA<pB, you'd expect B would hit the target before A does generally. So the probability P(nB>nA) should be different in the two cases, but your answer would say they're the same.

9. Jun 23, 2010

### SamTaylor

Oh i think i found the mistake!

$$\frac{P_a (1-P_b)}{1-(1-P_b)(1-P_a)}$$

Which looks right to me, because when Pa goes higher, P(nb>na) -> Pb

10. Jun 23, 2010

### vela

Staff Emeritus
Looks good!

The summation could be written

$$P(N_A>N_B) = \sum_{n_a=1}^\infty \sum_{n_b=n_a+1}^\infty P(n_a,n_b) = \sum_{n_a=1}^\infty P_a(1-P_a)^{n_a-1} \sum_{n_b=n_a+1}^\infty P_b(1-P_b)^{n_b-1}$$

The sum

$$\sum_{n_b=n_a+1}^\infty P_b(1-P_b)^{n_b-1}$$

is the probability P(NB>NA|N_a=nA) that B shoots more times than A does given that A hits on shot nA. This is the probability that hint 2 asked you to find. The other sum corresponds to hint 3, so you have in the end

$$P(N_A>N_B) = \sum_{n_a=1}^\infty P(N_B>N_A|N_a=n_A)P(N_A=n_A)$$