Statistics/Probability

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Peachyyy

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A sharpshooter hits a target with probability .75. Assuming indepence, find the probablities of getting:

a) a hit followed by two misses
b) two hits and a miss in any order

I got the answer correct for a. by multiplying .75 * .25 * .25. HOWEVER, i feel like that was just luck, because it doesn't make much sense that they would have to be in that order. I could have written .25 * .25 * .75 and gotten the same answer.

Could someone please help?

Georgia
 
let h be a hit
let m be a miss

if you fire 3 times the total probabil;ities are each equally likely:
shot probability
1 2 3
h h h .75^3
h h t .75^2 x .25
h t h .75^2 x .25
h t t .75 x .25^2
t h h .25 x .75^2
t h t .25^2 x .75
t t h .25^2 x .75
t t t .25^3

probability of 3 hits one case
probability of 2 hits and 1 miss in any order 3 cases sum for the answer
probability of 1 hit and 2 misses in any order 3
probability of 3 misses 1 case

probability of 2 hits and 1 miss in a specific order 1 case

does this help?
Arthur
 
Ok, still confused. I thought since the probablility that he would make a hit is .75 that there wouldn't be equally likely possibilities? I don't understand the difference (or how to get the difference) between one hit followed by two misses and two hits and a miss in any order. Why are they different?
 
I'm not sure where the confusion is.

There are three equally likely ways of one hit and two misses. Probability of each is .75*.25²

There are three equally likely ways of one miss and two hits. Probability of each is .75²*.25

When you say in any order you have to add the three together (Or multiply by three) then there are four possible outcomes.
3 hits 1*.75³
2 hits 3*.75²*.25
1 hit 3*.75*.25²
0 hits 1*.25³

Note: gathering terms of
(.75h+.25m)³ = (.75³)h³+3*(.75²)(.25)h²m+3*(.75)(.25²)hm²+(.25³)m³
 
Peachyyy said:
Ok, still confused. I thought since the probablility that he would make a hit is .75 that there wouldn't be equally likely possibilities? I don't understand the difference (or how to get the difference) between one hit followed by two misses and two hits and a miss in any order. Why are they different?
Click to expand...
You're not shooting, writing down the results, and then drawing random results form a hat. You are recording results in succession. In order to hit and then miss, you must not miss on the first shot. Did you build a tree?


Code: 
HIT (0.75)  - HIT (0.75)  - HIT (0.75)
                          - MISS (0.25)
            - MISS (0.25) - HIT (0.75)
                          - MISS (0.25)
MISS (0.25) - HIT (0.75)  - HIT (0.75)
                          - MISS (0.25)
            - MISS (0.25) - HIT (0.75)
                          - MISS (0.25)
Each branch of the tree has its own string of probabilities. Some branches have the same total outcomes, but in different orders.
 
tkhunny said:
Each branch of the tree has its own string of probabilities. Some branches have the same total outcomes, but in different orders.
Click to expand...

I think this is where my confusion lies. the first question asks for a specific order: H M M; I got this by multiplying .75*.25*.25. HOWEVER, since multiplication is commutitive, that truly could have been a hit and two misses in any order. (I hope this makes sense). The second part is asking for H H M in any order. What's the difference? Why doesn't .75*.75*.25 work for this one? It seems to me (from my reasoning above) that this would make more sense for any order than it would for a specific order.

I'm sorry guys. I feel like a total idiot, but something's just not clicking.
 
Hmmmm,
We have said it about every way I can think of. .75*.75*.25 IS the probability of HHM. When you add "in any order" you are adding to the choices that are possible. HMH and MHH are now included. .75*.25*.75 and .25*.75*.75 which, as you say are commutations and give the same probability for each of the three "in this orders". I don't see why this bothers you.
Think about coins.
The probability of flipping two heads or two tails is the same, 1/4.
Flipping heads-tails is also 1/4.
Flipping tails-heads is also 1/4.
One head and one tail "in any order" makes it 1/4 + 1/4 = 1/2.
Does that bother you too?

Different choices of any kind CAN have the same probability.
 
Ok, that gave me a lightbulb moment. Sorry, I don't know what my problem is. I think i was trying to make it a lot more difficult than it was. Thanks for your help.
 
Consider
.75*.25*.25 -- HMM
.25*.75*.25 -- MHM

Those are the same total outcome.
Those are not the same path on the tree.
 
I suggest you write out all possibilities {which I did to try to show you the difference between a hit and two misses and a hit miss miss

look at the combinations
h h h
h h m
h m h
h m m
m h h
m h m
m m h
m m m

to determine probability of 1 hit and two misse in any order we have
h m m OR m H M OR m m h when you say or you add
.75 [ .25]^2 + .75[.25]^2 + .75[.25]^2 = 9/64

9 times out of 64 sets of 3 shots you will get 1 hit and 2 misses

BUT a h m m in that order = 1 case .75[.25^2]= 3/64

3 times out of 64 set of 3 shots will you get a h m m in that order
Arthur
 
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