Probability of hitting a bull's eye

[JJBladester]

Homework Statement

I'm studying for the FE exam and it's been a while since I've done some of the more basic math.

One of the practice problems for the math section states:

"A marksman can hit a bull's-eye from 100 m with three out of every four shots. What is the probability that he will hit a bull's-eye with at least one of his next three shots?"

Homework Equations

I don't know of any.

The Attempt at a Solution

The solution has been given but I want to solve it differently. The book's solution states:

P(miss) = 1- P(hit) = 1 - 3/4 = 1/4

P(at least one) = 1 - P(none) = 1- ((P(miss) X P(miss) X P(miss)) = 1- (1/4)(1/4)(1/4) = 63/64

I want to solve the problem by calculating and then summing the three probabilities:

P{at least 1 hit in 3 shots} = P{1 hit in 3 shots} + P{2 hits in 3 shots} + P{3 hits in 3 shots}

So my problem is how to find the probability of these three scenarios. I tried mapping out what different combinations there could be and found 8. See below (1 = hit, 0 = miss, column = attempt #).

0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

I thought the answer would be:

P{1 hit in 3 shots} + P{2 hits in 3 shots} + P{3 hits in 3 shots} = 3/8 + 2/8 + 1/8 = 3/4

... but it's not. How do I factor the P(hit) = 3/4 into the above equation?

 

[cpscdave]

You need to assign the relative probabilities weightings to each of those 3 shot possible outcomes.
IE
0 0 0 would be (1/4)*(1/4)*(1/4)
0 0 1 would be (1/4)*(1/4)*(3/4)
etc

so you would end up with (1/4)*(1/4)*(3/4) * 3 + (1/4)*(3/4)*(3/4) * 3 + (3/4)(3/4)(3/4) = 63/64

 

[phinds]

Also, how do you get, using your method, P(2 hits in 3 shots) = 2/8 ?

What matters in my question is that is clearly shows that you do not attempt even the simplest double-check on your work, and that will cause you grief in ALL problems. From your table, there are 7 ways out of 8 that there is at least one hit, so it should have jumped out at you that an answer of 6/8 (using your method) could not be correct.

 

[JJBladester]

cpscdave: Thank you! It makes sense now.

phinds: It did jump out at me that my answer was not correct. That's why I posted for help on physicsforums. Intuitively, if the marksman can hit 3/4 of his shots, he should have way more than a 0.75 probability of hitting at least one out of three shots. This is why I knew I was traveling down the wrong road and why I asked for help.

 

[Ray Vickson]

cpscdave: Thank you! It makes sense now.

phinds: It did jump out at me that my answer was not correct. That's why I posted for help on physicsforums. Intuitively, if the marksman can hit 3/4 of his shots, he should have way more than a 0.75 probability of hitting at least one out of three shots. This is why I knew I was traveling down the wrong road and why I asked for help.

There are two ways of computing $P(A)$ for an event $A$: (i) directly; and (ii) indirectly, by using $P(A) = 1 - P(A^c)$, where the event $A^c$ is the complement of $A$. In your case, (ii) is a much faster and more efficient method than (i)---and you did it correctly---but if you have the luxury of time you can also try (i) (which was giving you trouble). On an exam, faster and easier is the way to go, unless you are specifically told otherwise.

 

[JJBladester]

Thanks Ray. According to my study materials and various study websites, the FE allows slightly less than three minutes per question so I will focus my efforts on time-saving advice like what you mentioned. I couldn't have done this in any less than 10 minutes without a shortcut (using the compliment of A).