Binomials (1 Viewer)

[Dreamerish*~]

On the average, a typist has to correct one word in 800 words. Assuming that a page contains 200 words, find the probability of more than one correction per page.

Answer: 0.0625.

Thank you.

 

[who_loves_maths]

hi Dreamerish,

when you say "find the probability of more than one correction per page"

are you saying that there is more than just ONE page involved?

if so, how many pages altogether? (4?)

 

[MarsBarz]

200 words

Pmistake = 1/800 * 200 = 0.25
Pcorrect = 799/800 * 200 = 199.75

mistake --> correction

P0correction = 200C200 * (199.75^200) * [0.25^(200-200)]
P0correction = (199.75^200)

P1correction = 200C199 * (199.75^199) * [0.25^(200-199)]
P1correction = 50 * (199.75^199)

Pmorethan1correction = 1 - P0correction - P1correction
Pmorethan1correction = 1 - (199.75^200) - 50 * (199.75^199)
Pmorethan1correction = 1 - (199.75^199)(199.75 - 50)
Pmorethan1correction = 1 - 149.75(199.75^199)

It can't possibly have such high powers which means that I have made a mistake somewhere. This is my first time doing binomial probability so I guess I have an excuse.

 

[Dreamerish*~]

hi Dreamerish,

when you say "find the probability of more than one correction per page"

are you saying that there is more than just ONE page involved?

if so, how many pages altogether? (4?)

Hello.

Nup, I copied the question exactly as it is. It doesn't state the number of pages.

Maybe it means just one page then?

 

[grendel]

assumption: probability that any particular word needs correcting is 1/800.

tf

p(more than one correction per page)=1-(p(0 corrections)+p(1 correction))

=1-(200C0 (799/800)^200 + 200C1 (1/800)(799/800)^199)
= 0.026

which is not the answer.

i think the method is correct but the initial assumption is wrong.

can somebody suggest another probability?

i've encountered similar questions in cambridge with little success in solving them.

is this question from cambridge??

 

[Dreamerish*~]

assumption: probability that any particular word needs correcting is 1/800.

tf

p(more than one correction per page)=1-(p(0 corrections)+p(1 correction))

=1-(200C0 (799/800)^200 + 200C1 (1/800)(799/800)^199)
= 0.026

which is not the answer.

i think the method is correct but the initial assumption is wrong.

can somebody suggest another probability?

i've encountered similar questions in cambridge with little success in solving them.

is this question from cambridge??

MMHMM! That is exactly what I did, and that's the answer I got, except I rounded it off to 4 significant figures.

I'm thinking maybe that's the right answer. So far 4 people (including you) have come up with the exact same answer. The question was from Fitzpatrick, which is sometimes wrong. However, the person who used my book last year and took the effort to circle every single challenging question, or questions where the answer in the book was wrong, did not circle that one.

 

[grendel]

the answer in the back of fitzpatrick is wrong.

had a look at the official solutions and the one i posted earlier is correct =)

 

[Roguedeth]

uh huh i got .0264076039

X = Number of bad words
P=1/800
N=200

they want Pr(x>1) = 1 - Pr (x<1)

so 1-binomcdf(200,1/800,1)

 

[rama_v]

the answer in the back of fitzpatrick is wrong.

had a look at the official solutions and the one i posted earlier is correct =)

Yeah I remember this question, the solution in teh back is definitley wrong

 

[Dreamerish*~]

Another question:

A marksman finds that on the 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:

(i) He hits the target each time

(ii) He scores at least 2 bull's eyes

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

I've got the first and second part. For the third, I have 281/1250 but the answer is 177/1250.

 

[who_loves_maths]

hi Dreamerish,

there are a number of ways to do part (iii), and i suspect you want the binomial version... but since i'm in a kind of rush tonight, i'll just do it using the "normal" way - you can figure out the binomial expansion yourself (not very hard), so sorry about this ; anyways:

4 rounds:

P(2 bull's eyes & all targets) = P(all targets) - P(0 bull's eye & all targets) - P(1 bull's eye & all targets)

Now,
P(hit) = 9/10 = P(hit & no bull's eye) + P(hit & bull's eye) = P(hit & no bull's eye) + 1/5
---> ie. P(hit & no bull's eye) = 9/10 - 1/5 = 7/10


Therefore:

P(2 bull's eyes & all targets) = (9/10)^4 - (7/10)^4 - (4C1).(7/10)^3.(1/5) = 177/1250 = 0.1416


hope that helps you

P.S. binomial expansion is of the form (in this case): ((7/10) + (2/10))^4

 

[Dreamerish*~]

Thank you so much!