Confusign probability

**jacs** · May 21st 2006, 06:15 AM

A marksman finds that on average he hits the target 9 times out of every 10 and scores a bull's eye on the average once every 5 rounds. He fires 4 rounds. What is the probability that:

He scores at least 2 bull's eyes and he has hit the target on each of the four rounds?

I have found that the prob he scores at least 2 bulls eyes to be 0.1808
and the prob of hittign the target all four times to be 0.6561

what i am not sure how to do is combine them

**CaptainBlack** · May 21st 2006, 07:01 AM

Originally Posted by jacs

A marksman finds that on average he hits the target 9 times out of every 10 and scores a bull's eye on the average once every 5 rounds. He fires 4 rounds. What is the probability that:

He scores at least 2 bull's eyes and he has hit the target on each of the four rounds?

I have found that the prob he scores at least 2 bulls eyes to be 0.1808
and the prob of hittign the target all four times to be 0.6561

what i am not sure how to do is combine them

Probability of a b-eye given that the target is hit is $2/9$ ,
so the probability of 2 b-eyes in 4 hits is:

$ P(2 \mbox{b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 \approx 0.1792 $

The probability of four hits from four shots is $0.9^4 \approx 0.6561$ .

So the probability of 2 b-eyes and four hits from four shots is:

$ P(2 \mbox{b-eyes and 4 hits from 4 shots})=$ $P(\mbox{2 b-eyes from 4 hits})$ $P(\mbox{4 hits from 4 shots})$

$ \approx 0.1792 \times 0.6561 \approx 0.1176 $

RonL

**jacs** · May 21st 2006, 05:47 PM

Originally Posted by CaptainBlack

Probability of a b-eye given that the target is hit is $2/9$ ,
so the probability of 2 b-eyes in 4 hits is:

$ P(2 \mbox{b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 \approx 0.1792 $

The probability of four hits from four shots is $0.9^4 \approx 0.6561$ .

So the probability of 2 b-eyes and four hits from four shots is:

$ P(2 \mbox{b-eyes and 4 hits from 4 shots})=$ $P(\mbox{2 b-eyes from 4 hits})$ $P(\mbox{4 hits from 4 shots})$

$ \approx 0.1792 \times 0.6561 \approx 0.1176 $

RonL

Thanks for that Captain Black, but does it change the probability if it is AT LEAST two bulls-eyes?

i got the prob for that to be

$ P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 $

$+{4 \choose 3} (2/9)^3 (7/9)+{4 \choose 4} (2/9)^4\approx 0.1808 $

then if i just multiply that with the 0.5661 i get 0.1186 and the text is claiming it is 0.1416

What i was wondering too, was if you get a bullseye, isnt it implicit that the target is hit? Or am i over thinking this?

thanks

**CaptainBlack** · May 21st 2006, 07:36 PM

Originally Posted by jacs

Thanks for that Captain Black, but does it change the probability if it is AT LEAST two bulls-eyes?

i got the prob for that to be

$ P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (2/9)^2 (7/9)^2 $

$+{4 \choose 3} (2/9)^3 (7/9)+{4 \choose 4} (2/9)^4\approx 0.1808 $

Check you arithmetic, I make this 0.2158

then if i just multiply that with the 0.5661 i get 0.1186 and the text is claiming it is 0.1416

You should be multiplying by 0.6561 here.

What i was wondering too, was if you get a bullseye, isnt it implicit that the target is hit? Or am i over thinking this?

Yes, but you will need to think about the calculation quite carefully to get
the right answer.

RonL

**jacs** · May 21st 2006, 08:32 PM

Originally Posted by CaptainBlack

Check you arithmetic, I make this 0.2158

You should be multiplying by 0.6561 here.

Yes, but you will need to think about the calculation quite carefully to get
the right answer.

RonL

yes, my bad, I meant to type 0.6561, got a bit of number dyslexia going there and apparently i cannot press calculator buttons either...grrr....duh to me.... so i had it right all along and just can't calculate properly.

thanks for that then and sorry for wasting your time.... next time instead of spending ages on the theory, i will double check my button pressing

**jacs** · May 21st 2006, 08:58 PM

I think we are looking at the wrong probablity and I just went along for the ride. For at least two two bulls-eyes we must use the bulls-eyes odds which is 1/5, the hitting the target was 2/9.

i got the prob for that to be

$ P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (1/5)^2 (4/5)^2 $

$+{4 \choose 3} (1/5)^3 (4/5)+{4 \choose 4} (1/5)^4\approx 0.1808 $

then if i just multiply 0.1808 with the 0.6561 i get 0.1186 and the text is claiming it is 0.1416

this whole question has just confused me completely and i dont know if i am just misinterpreting the question now

a) P(hits all four times) = 0.6561 which was correct
b) P(hits at least 2 bulls-eyes) = 0.1808 which was correct

c) P(scores at least 2 bulls-eyes and hits all four shots)

so i figured multiplying them woudl yeild the correct result
but it doesnt, 0.6561 x 0.1808 = 0.1186

your method gets that result, 0.6561 x 0.2158 = 0.1416
but then it would invalidate the previous result in b, wouldnt it?

or has the text just screwed up???

**CaptainBlack** · May 22nd 2006, 08:52 AM

Originally Posted by jacs

I think we are looking at the wrong probablity and I just went along for the ride. For at least two two bulls-eyes we must use the bulls-eyes odds which is 1/5, the hitting the target was 2/9.

i got the prob for that to be

$ P(2 \mbox{at least 2 b-eyes from 4 hits})={4 \choose 2} (1/5)^2 (4/5)^2 $

$+{4 \choose 3} (1/5)^3 (4/5)+{4 \choose 4} (1/5)^4\approx 0.1808 $

then if i just multiply 0.1808 with the 0.6561 i get 0.1186 and the text is claiming it is 0.1416

In order for multiplication to be valid here the two events represented by
the probabilities have to be independent, but here they are not, because
the b-eyes imply a hit.

My approach using the conditional probability of a b-eye gets rid of this
problem in a systematic manner using:

$ p(a \mbox{ and } b)=p(a \mbox{ given }b).p(b) $

RonL

**jacs** · May 22nd 2006, 05:38 PM

Originally Posted by CaptainBlack

In order for multiplication to be valid here the two events represented by
the probabilities have to be independent, but here they are not, because
the b-eyes imply a hit.

My approach using the conditional probability of a b-eye gets rid of this
problem in a systematic manner using:

$ P(a \mbox{ and } b)=p(a \mbox{ given }b).p(b) $

RonL

I see, we haven't covered conditional probabilities yet. But i do understand what you mean about the independence, I knew there was some problem there, but i just had no idea what to do about it.

thank you for all your help
jacs

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