Binomial Probability (1 Viewer)

[vds700]

from Ex 29b) fitzpatrick 3 unit book.

20. On average a typist has to correct 1 word in 800 words. Assuming that a page contains 200 words, find the probability of more than 1 correction per page.

22. Four families each have 4 children. What is the probability that:
(i) At least one of these families has 2 boys and 2 girls.
(ii) each family has at least 1 boy?

 

[lolokay]

from Ex 29b) fitzpatrick 3 unit book.

20. On average a typist has to correct 1 word in 800 words. Assuming that a page contains 200 words, find the probability of more than 1 correction per page.

22. Four families each have 4 children. What is the probability that:
(i) At least one of these families has 2 boys and 2 girls.
(ii) each family has at least 1 boy?

not sure about the first one. I would think it's 1 - (799/800)^200 - 200(799/800)^199 * (1/800) = 0.0264...


second one: combinations of 2/4 boys is 4c2 = 6. Total number of combinations = 2^4 = 16, so 10/16 chance of not having 2 boys and 2 girls. Chance of all 4 families not having 2 boys, 2 girls is therefore (10/16)^4. Chance of at least one family having 2 boys 2 girls is 1 - (10/16)^4

for each family, there's a 1/16 chance of not having a boy, so a 15/16 chance of having a boy. So the chance of all 4 families having a boy is (15/16)^4

 

[lyounamu]

from Ex 29b) fitzpatrick 3 unit book.

20. On average a typist has to correct 1 word in 800 words. Assuming that a page contains 200 words, find the probability of more than 1 correction per page.

22. Four families each have 4 children. What is the probability that:
(i) At least one of these families has 2 boys and 2 girls.
(ii) each family has at least 1 boy?

1) 800 words = 4 pages

P(more than 1 correction per page) = 1 - (4C0 . 799/800^4 . 1/800^0 + 4C1 . 799/800^3 . 1/800^1) = 9.35 . 10^-6.

Um... the answer is definitely really small. But mine is too small.

2i) P(2 boys, 2 girls) = 4C2/2^4 = 3/8
P (NOT 2 boys, 2 girls) = 5/8

P(At least one family) = 1 - 4CO . (5/8)^4 . (3/8)^0 = 3471/4096

2ii) P(not having a boy) = 4CO / 2^4 = 1/16
P (having a boy) = 15/16

P(at least one) = 4C4 . (1/16)^0 . (15/16)^4 = 0.77247... = 0.772

 

[lolokay]

1) 800 words = 4 pages

P(more than 1 correction per page) = 1 - (4C0 . 799/800^4 . 1/800^3 + 4C1 . 799/800^3 . 1/800^1) = 9.35 . 10^-6.

Um... the answer is definitely really small. But mine is too small.

I don't get what you did there, could you please explain? and the number of pages in 800 words doesn't matter does it?

btw, I put my working into the calculator wrongly the first time for that question, and we got the same answer for the other 2 questions

 

[vds700]

1) 800 words = 4 pages

P(more than 1 correction per page) = 1 - (4C0 . 799/800^4 . 1/800^3 + 4C1 . 799/800^3 . 1/800^1) = 9.35 . 10^-6.

Um... the answer is definitely really small. But mine is too small.

2i) P(2 boys, 2 girls) = 4C2/2^4 = 3/8
P (NOT 2 boys, 2 girls) = 5/8

P(At least one family) = 1 - 4CO . (5/8)^4 . (3/8)^0 = 3471/4096

2ii) P(not having a boy) = 4CO / 2^4 = 1/16
P (having a boy) = 15/16

P(at least one) = 4C4 . (1/16)^0 . (15/16)^4 = 0.77247... = 0.772

Namu u sure the first one is right? The answer in the book is 0.0625, though its possible this is wrong. The bold part doesn't look right to me.

Where did u get the 4 from, isn't it 1 page?

What i did was work with the no. of words (200)
P(> 1 correction per 200 words) = 1 - [P(0 corrections) + P(1 correction)]
=1 - [(799/800)^200 + 200C1(1/800)(799/800)^199]
=0.02639

 

[lyounamu]

Namu u sure the first one is right? The answer in the book is 0.0625, though its possible this is wrong. The bold part doesn't look right to me.

Where did u get the 4 from, isn't it 1 page?

What i did was work with the no. of words (200)
P(> 1 correction per 200 words) = 1 - [P(0 corrections) + P(1 correction)]
=1 - [(799/800)^200 + 200C1(1/800)(799/800)^199]
=0.02639

I knew my answer was different from the answer and I reckon I got it wrong.

Well. There were 800 words. 800 words = 4 pages. That's how I deduced it.

 

[lyounamu]

I don't get what you did there, could you please explain? and the number of pages in 800 words doesn't matter does it?

btw, I put my working into the calculator wrongly the first time for that question, and we got the same answer for the other 2 questions

Um, the bolded line by vds is my mistake. It should be (1/800)^0.

P(getting MORE THAN one correction) = 1 - (no correction + one correction)

I think my working out is wrong. What I did was, probability of getting correction = 1/800 and not getting = 799/800.

Since 800 words = 4 pages, what I did from here is:

P(getting more than once correction) = 1 - (4C0 . 799/800^4 . 1/800^0 + 4C1 . 799/800^3 . 1/800^1) = my answer.

 

[vds700]

Um, the bolded line by vds is my mistake. It should be (1/800)^0.

P(getting MORE THAN one correction) = 1 - (no correction + one correction)

I think my working out is wrong. What I did was, probability of getting correction = 1/800 and not getting = 799/800.

Since 800 words = 4 pages, what I did from here is:

P(getting more than once correction) = 1 - (4C0 . 799/800^4 . 1/800^0 + 4C1 . 799/800^3 . 1/800^1) = my answer.

yeah but they want the probabilility for more than 1 correction per page, not for 4 pages.

 

[lyounamu]

yeah but they want the probabilility for more than 1 correction per page, not for 4 pages.

Yeah. I did use one page thing. I got really weird answer.

But don't they want the probability in which more than one correction occurs? Which means that we cannot use the 1C1. Instead we shoud use NC1 since we don't know the numbers of papers. But if I do that, I get nowhere.

 

[lolokay]

Namu u sure the first one is right? The answer in the book is 0.0625, though its possible this is wrong. The bold part doesn't look right to me.

Where did u get the 4 from, isn't it 1 page?

What i did was work with the no. of words (200)
P(> 1 correction per 200 words) = 1 - [P(0 corrections) + P(1 correction)]
=1 - [(799/800)^200 + 200C1(1/800)(799/800)^199]
=0.02639

that's what my working is too. I don't see anything wrong with it

 

[lyounamu]

that's what my working is too. I don't see anything wrong with it

Well I tried every possible method basically. I originally tried vds and yours and I tried for the one page thing. I tried the N thing. I basically tried all the methods that I could think of. If the answer is not 0.02.. or 9.5 . 10^-6, I do not know how they got the answer as 0.0625.

I think there is something tricky about the wording of the question.

 

[lyounamu]

Our answers may not be wrong actually. That is because the solutions of the Fitzpatrick can be wrong in my book. My book is an old edition. The answer might be different on the newer edition.

Can anyone care to post solution? (from newer edition).

 

[vds700]

Well I tried every possible method basically. I originally tried vds and yours and I tried for the one page thing. I tried the N thing. I basically tried all the methods that I could think of. If the answer is not 0.02.. or 9.5 . 10^-6, I do not know how they got the answer as 0.0625.

I think there is something tricky about the wording of the question.

Namu did u or someone ask Mr Haman to do this question? If so, what did he get?

 

[lyounamu]

Namu did u or someone ask Mr Haman to do this question? If so, what did he get?

He wouldn't get it. And no, I didn't ask him. I will make sure I ask him.

I reckon it will be Justin again who solves this.

 

[vds700]

Our answers may not be wrong actually. That is because the solutions of the Fitzpatrick can be wrong in my book. My book is an old edition. The answer might be different on the newer edition.

Can anyone care to post solution? (from newer edition).

i seriously doubt its changed. My copy was new in 2006. I'd be surprised if Fitzpatrick is still alive, he wrote it in 1984 lol. And apparantly he got his students to do the answers, which would explain why there are mistakes.

 

[lyounamu]

i seriously doubt its changed. My copy was new in 2006. I'd be surprised if Fitzpatrick is still alive, he wrote it in 1984 lol. And apparantly he got his students to do the answers, which would explain why there are mistakes.

They still bring out new editions though. Justin has got a textbook that has got different solution to mine. He has got more accurate answers. But if your solution is 2006 one, I think that would be quite accurate.

Yeah, I heard that the students did that. Then how did they get it?

 

[watatank]

What i did was work with the no. of words (200)
P(> 1 correction per 200 words) = 1 - [P(0 corrections) + P(1 correction)]
=1 - [(799/800)^200 + 200C1(1/800)(799/800)^199]
=0.02639

The books wrong, you have done the question right.

 

[lyounamu]

The books wrong, you have done the question right.

Then how do you know that?

 

[lyounamu]

Then let's just say that the solution on the textbook is wrong. Sounds fair, yeah???

 

[vds700]

Then let's just say that the solution on the textbook is wrong. Sounds fair, yeah???

yeah i reckon so