Binomial Probability Q - Annoying the Heck out of me! (1 Viewer)

[lookoutastroboy]

Hey guys,

This question has been bugging me for ages! Its from 3U Fitzy book so if anyone can solve it, it'd be much appreciated.

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

iii. he scores at least 2 bull's eyes and he has hit the target on each of the 4 rounds?

Answer: 177/1250





Cheers,
l.a.

 

[maths_is_easy5]

that is PISS easy.

You fail

 

[michaeljennings]

that is PISS easy.

You fail

why do you run away from the threads were people are owning you. It makes it hard for BOSers to continue laughing at you cos they gotta follow about 50 threads to read all the witty insults people dish out to you

 

[lookoutastroboy]

thanks bro

 

[lookoutastroboy]

michael, can u solve the problem?

 

[michaeljennings]

michael, can u solve the problem?

mate sorry i gotta get to sleep now was meant to go an hour ago but dw someone will post here soon with some helpful info, ill check this thread tomorrow just incase no one has

 

[pwoh]

This is what I got

Probability of hitting target on all 4 rounds: (9/10)^4

Probability of 0 bullseye = (4/5)^4
Probability of 1 bullseye = 4 (1/5)(4/5)^3 = (4/5)^4

Probability of 2+ bullseye = 1 - 2*(4/5)^4

Total = (1 - 2*(4/5)^4) * ( (9/10)^4)

Doesn't match the answer though =| Were there any previous sub-questions?

 

[taeyang]

I got 0.1438........ hmmmmm

 

[hscishard]

9/10 shots make it. 1/5 shots are bullseyes, therefore 2/10 are expected to be bullseyes.
Of the shots that make it in 10 shots, 2/9 are expected to be bullseye and 7/9 are not.

Without me writing up a whole heap, your x and y or your p and q (whatever you use to represent chances of success or failure) would be (9/10)(2/9) and (9/10)(7/9)

 

[someth1ng]

Are you sure that's the right answer? Fitzpatrick are known for alot of errors.

 

[deterministic]

hschard is on the right track.

P(hitting bulleyes)=1/5=2/10
P(hitting target)=9/10
So P(hitting target but not hitting bulleyes)=9/10-2/10=7/10

So:
P(hitting 2 bulleyes and 2 targets(but not bulleyes))=(7/10)^2*(2/10)^2*4C2
P(hitting 3 bulleyes and 1 targets(but not bulleyes))=(7/10)^1*(2/10)^3*4C1
P(hitting 4 bulleyes)=(2/10)^4

Then add the above probabilities together to get the answer

 

[pwoh]

hschard is on the right track.

P(hitting bulleyes)=1/5=2/10
P(hitting target)=9/10
So P(hitting target but not hitting bulleyes)=9/10-2/10=7/10

So:
P(hitting 2 bulleyes and 2 targets(but not bulleyes))=(7/10)^2*(2/10)^2*4C2
P(hitting 3 bulleyes and 1 targets(but not bulleyes))=(7/10)^1*(2/10)^3*4C1
P(hitting 4 bulleyes)=(2/10)^4

Then add the above probabilities together to get the answer

D'oh, I just realised that a bullseye is on the target...

 

[hscishard]

How strange, initially I used the 7/10 way but never arrived at the answer.
And now after looking at deterministic's answer, 9/10 x 7/9 = 7/10. Mindfuck lol

 

[hscishard]

u r shit son

Long time no talk, bro