Topic: URGENT MATHS HELP NEEDED!

[OohCampione]

Speke to Bootle

Ten past the hour, half past the hour and ten to the hour.

Not to be mixed up with the 81A

[blert596]

Speke to Bootle

Ten past the hour, half past the hour and ten to the hour.

Not to be mixed up with the 81A

Used to be the 81C from the match back to canny for me.

If I worked it out right.



Oh no i didnt. I actually lived in number 81, and got the 12A. I think thats right.

[ollick]

Think it's 81.

[rakey_lfc]

Think it's 81.

hmmm, I think you're right.

[Doc Evil]

You just wait untill the 11 times table comes home.

Love the 11 times table - not a patch on the 9 times table though.

[Doc Evil]

hmmm, I think you're right.

He's wrong it's 81

[callanlfc5times]

Bump got this assignment due in a couple of days and need to know about binomial distribution? Got to do this question I have the answers just don't know how to work them out. Yes I am trying to get someone to do my assignment for me haha. But seriously if anyone could help me out would be great I really need to learn this and google throws out the most complicated formulas which is really confusing

The publishers of a magazine available only through direct subscription noted that exactly 60% of its subscribers are male. Four of the subscribers are randomly chosen to receive prizes. What are the probabilities that:
      a).   no males receive a prize,
      b).   exactly 2 males receive prizes,
      c).   at least 3 males receive prizes?

(answers a). 0.0256, b). 0.3456, c). 0.4752)

[Stevie93]

nCr x (1 - p)^(n - r) x p^r

a) For p = 0,   4C0 x (1 - 0.6)^(4 - 0) x 0.6^0 = 0.0256

b) For p = 2,   4C2 x (1 - 0.6)^(4 - 2) x 0.6^2 = 0.3456

c) 1 - ((p = 0) + (p=1) + (p = 2))
   
    For p = 1,  4C1 x (1 - 0.6)^(4 - 1) x 0.6^1 = 0.1536
 
    So 1 - (0.0256 + 0.3456 + 0.1536) = 0.4752

Does that help?
It would be far easier if you had a table of cumulative binomial probabilities. Do you?

[callanlfc5times]

No don't have one of them, I think we are just meant to work them out with the use of a formula  Cheers for that mate I'll have a look through what you have done and try to decipher what I can from it and hopefully it will be enough to help me with the other questions.

[SMD]

http://academic.regis.edu/jwise/Table%20%204%20%20AND%205.pdf

[callanlfc5times]

I give up!

[Ferg]

The answer is 81 mate, in case you were wondering.

No it's not. It's 42. That's the answer to everything.

[callanlfc5times]

The answer is 81 mate, in case you were wondering.

I know the answers, the answers are already given. The marks are for showing how to work the questions out.

[Ken-Obi]

It's 81.

[blert596]

how many choices are there?


Its C.

[callanlfc5times]

Binomial distribution question? Anyone fancy doing this one and all?

Out of 800 families with 5 children how many would you expect to have:
a). 3 boys,
b). 5 girls,
c). either 2 or 3 boys?
(250, 25, 500)

[blurred]

Are you going to take a laptop into the exam and ask RAWK to get your degree for you, too?

[Drippy Dick]

Binomial distribution question? Anyone fancy doing this one and all?

Out of 800 families with 5 children how many would you expect to have:
a). 3 boys,
b). 5 girls,
c). either 2 or 3 boys?
(250, 25, 500)

This is really probability cum binomial distribution.
Total possiblities is 2^5 ie 32

a) No. of arrrangements of 3 boys and 2 girls (since 3 boys will imply other 2 are girls) = 5!/(3! * 2!) = 10
Thus answer will be 800*10/32 = 250

b) No. or arrangements of 5 girls = 1 (GGGGG)
Thus answer will be 800 * 1/32 = 25

c) Total expectation of 3 boys = 250 which is same as expectation of 3 girls
3 girls implies 2 boys
Thus expectation of 3 boys or 2 boys is same as expectation of 3 boys and 3 girls = 250 + 250 = 500

Sorted

! imples factorial

[callanlfc5times]

Cheers I feel kind of cheeky but what the hell it's getting my work done and I don't have a clue what I'm doing, anyone fancy doing this? Binomial distribution again.

What is the probability of getting 9 exactly once in 3 throws of a pair of fair dice?

[callanlfc5times]

A computer generates single-digit random numbers from 0 to 9 inclusive. Each digit is equally likely to be generated. If 50 such digits are generated what are the probabilities of obtaining:
a). 5 or more twos,
b). 4 or less threes,
c). between 6 and 10 (inclusive) sevens?
(0.5688, 0.4312, 0.3745)

[Drippy Dick]

Cheers I feel kind of cheeky but what the hell it's getting my work done and I don't have a clue what I'm doing, anyone fancy doing this? Binomial distribution again.

What is the probability of getting 9 exactly once in 3 throws of a pair of fair dice?

Total possible ways to throw a pair of dice are 6*6 ie 36
No. of ways to get 9 is 4+5, 5+4, 3+6 and 6+3 ie 4 ways.

So you can get 9 in 4/36 ie 1/9 ways every time you throw a pair of dice. This implies you can get numbers besides 9 8 out of 9 times you throw ie probability is 8/9 for something besides 9.
Now you need 9 only once in 3 throws.

ie 9 on the first throw and something else on 2nd and 3rd throws. Probability is (1/9)* (8/9) * (8/9) = 64/729
or 9 on second throw and something else on 1st and 3rd. Probability is (8/9)* (1/9) * (8/9) = 64/729
Similarly 9 only on third throw is also 64/729

Making total probability (64/729) + (64/729) + (64/729) = 192/729

Note: Probability of event A OR event B = sum of the two individual probabilities
Probability of event A AND event B = Product of the two individual probabilities


Will get to the enxt one if I can be arsed and have more time later 

[callanlfc5times]

Total possible ways to throw a pair of dice are 6*6 ie 36
No. of ways to get 9 is 4+5, 5+4, 3+6 and 6+3 ie 4 ways.

So you can get 9 in 4/36 ie 1/9 ways every time you throw a pair of dice. This implies you can get numbers besides 9 8 out of 9 times you throw ie probability is 8/9 for something besides 9.
Now you need 9 only once in 3 throws.

ie 9 on the first throw and something else on 2nd and 3rd throws. Probability is (1/9)* (8/9) * (8/9) = 64/729
or 9 on second throw and something else on 1st and 3rd. Probability is (8/9)* (1/9) * (8/9) = 64/729
Similarly 9 only on third throw is also 64/729

Making total probability (64/729) + (64/729) + (64/729) = 192/729

Note: Probability of event A OR event B = sum of the two individual probabilities
Probability of event A AND event B = Product of the two individual probabilities


Will get to the enxt one if I can be arsed and have more time later 

Cheers mate, much apreciated. I'd be fucked without this.

[Stevie93]

Mate just learn it, it's not hard. What is this, AS level S1?

[callanlfc5times]

Mate just learn it, it's not hard. What is this, AS level S1?

Don't have a clue, it's uni work which makes it even more fucking annoying because I'm doing a computing course and I don't see how it fits in with computers in the slightest. I disagree it is hard I haven't done any kind of maths for two years and only just managed to scrape a C at GCSE and that was only because I needed only 17 marks on each paper.

The formulas are mind boggling I can't make any sense of them.

[AndyInVA]

nCr x (1 - p)^(n - r) x p^r

a) For p = 0,   4C0 x (1 - 0.6)^(4 - 0) x 0.6^0 = 0.0256

b) For p = 2,   4C2 x (1 - 0.6)^(4 - 2) x 0.6^2 = 0.3456

c) 1 - ((p = 0) + (p=1) + (p = 2))
   
    For p = 1,  4C1 x (1 - 0.6)^(4 - 1) x 0.6^1 = 0.1536
 
    So 1 - (0.0256 + 0.3456 + 0.1536) = 0.4752

Does that help?
It would be far easier if you had a table of cumulative binomial probabilities. Do you?

this is an awesome answer posted within a matter of minutes

RAWK is amazing

[AndyInVA]

Are you going to take a laptop into the exam and ask RAWK to get your degree for you, too?

hahhahaha

[callanlfc5times]

Na no exam for this module.   

[blurred]

Na no exam for this module.   

Ah, that's good. Just the sense of having earned your degree then. Sound.

[callanlfc5times]

I'm not complaining it's not me who decides if there should be an exam or not.

[SMD]

Don't have a clue, it's uni work which makes it even more fucking annoying because I'm doing a computing course and I don't see how it fits in with computers in the slightest. I disagree it is hard I haven't done any kind of maths for two years and only just managed to scrape a C at GCSE and that was only because I needed only 17 marks on each paper.

The formulas are mind boggling I can't make any sense of them.

Think you should change your degree.

[blurred]

See, now, I was being nice and letting that one slide...

[SMD]

[blert596]

See, now, I was being nice and letting that one slide...

What were the chances of that happening?

[surfer. Fuck you generator.]

A computer generates single-digit random numbers from 0 to 9 inclusive. Each digit is equally likely to be generated. If 50 such digits are generated what are the probabilities of obtaining:
a). 5 or more twos,
b). 4 or less threes,
c). between 6 and 10 (inclusive) sevens?
(0.5688, 0.4312, 0.3745)

The script for the question is straightforward, but the calculation is a bit long. Essentially:

n = number of trials = 50
x = number of successful events (dependent on question)
p = probability of success in one trial = 1/10 for any digit as stated in the question.

Again, the calculation formula is: nCx times P to the power x times  (1 - P) to the power( n - x )

Don't know how to superscript on here, so here's a link that explains it well. This is the essential bit. http://stattrek.com/Lesson2/Binomial.aspx

The nCx is the formula for combinations.

a) 5 or more '2' (5 is inclusive)
The method to calculate would be 1 -  p[0 number of '2' + 1 number of '2' + 2 no of '2' + 3 no of '2' + 4 no of 2]
so, 1 - 0.4312 (4 decimal places) = 0.5688

b) The answer's already in 'a' : The calculation in the brackets was for the quantity 4 or less of any digit: 0.4312 (4 decimal places)

c)I did it like this:
P(less than 11 '7') - P(less than 6 '7')
   0.9906...  - 0.6161...
0.3745 (4 decimal places)

I used the binomial calculator on that site, but a scientific calculator is necessary. Maybe someone else here can provide a simpler way of solving it.

[SP]

No it's not. It's 42. That's the answer to everything.

42 is the answer to Life, the Universe and Everything. Bonjela is the answer to everything.

[callanlfc5times]

Think you should change your degree.

Na, will just drop this module at the end of the year.

[SMD]

This thread is brilliant.

If you think this is great, check this out.

[kavah]

can someone show me how to solve this type of problem please.
ta

solve W?

0.75 = (0.5xW) + [0.9 (1-W)]

[SkinHimHesShite]

multiply the brackets out then solve. when you throw a number over the other side of the "=" its sign changes.

0.75 = 0.5W + 0.9 - 0.9W
- 0.15 = -0.4 W
W = 0.15/0.4 = 0.375

[kavah]

multiply the brackets out then solve. when you throw a number over the other side of the "=" its sign changes.


0.75 = 0.5W + 0.9 - 0.9W
- 0.15 = -0.4 W
W = 0.15/0.4 = 0.375

thanks mate.

multiply the brackets out, genius! 
I'm studying for a finance exam (for non-financial people) and the rest of my class are about 20 years younger than me and about 20 times better at maths so I was too slow in my class note making.

much appreciated