Help with Binomial Maths? xD

Cookeh · 11 Feb 2010 at 20:12

Anyone one able to work out the following binomial distribution question? Its had me stumped for ages :confused:

"Some eggs sold in a store are packed in boxes of 10. For any egg, the probability that it is is cracked is 0.05, independently of all other eggs. Calculate the expected number of cracked eggs in a box."

I got as far as C~B(10, 0.05) where C = cracked eggs. But stumped from there

Please dont spam saying Im dense or anything similar

Thanks in advance,
Cookeh

AJK · 11 Feb 2010 at 20:14

Taken literally the answer is 10, since the question doesn't ask for the expected number of uncracked eggs in a box.

Cookeh · 11 Feb 2010 at 20:15

AJK said:
Taken literally the answer is 10, since the question doesn't ask for the expected number of uncracked eggs in a box.

Sorreh! I really am dense xD
Fixed for correctness

G-Dubs · 11 Feb 2010 at 20:29

Never heard of binomunal maths, so guess this is "probably" wrong, the probability of a cracked egg is 5% which would be easy if there were 20 eggs in a basket (1 egg cracked)

So number of eggs cracked = 1/2 of an egg

That's about as far as I can get yer buddy.

Biohazard · 11 Feb 2010 at 20:30

that would be an ecumenical matter

Cookeh · 11 Feb 2010 at 20:33

G-Dubs said:
Never heard of binomunal maths, so guess this is "probably" wrong, the probability of a cracked egg is 5% which would be easy if there were 20 eggs in a basket (1 egg cracked)

So number of eggs cracked = 1/2 of an egg

That's about as far as I can get yer buddy.

I arrived there.
But not 100% sure of it.
And its A2 level stuff, think its fairly new to the syllabus?

Pudney · 11 Feb 2010 at 20:40

I did binomial maths in my A-Level maths. That was ages ago though, I do remember it was pretty easy to me (but Stats was my best area of maths by far). However, this has been entirely unhelpful to you, apologies!

Mansize_tissue · 11 Feb 2010 at 20:42

Finding the expectation when dealing with the binomial distribution simply involves multiplying the number of things by the probability

E(X) = np = 10*0.05 = 0.5

Cookeh · 11 Feb 2010 at 20:44

Pudney@work said:
I did binomial maths in my A-Level maths. That was ages ago though, I do remember it was pretty easy to me (but Stats was my best area of maths by far). However, this has been entirely unhelpful to you, apologies!

lol, no problem at all xD
This is oddly the only thing I have yet to struggle with at maths /:

Mansize_tissue said:
Finding the expectation when dealing with the binomial distribution simply involves multiplying the number of things by the probability

E(X) = np = 10*0.05 = 0.5

Awesome, thanks. Didnt think it would be that easy

Pudney · 11 Feb 2010 at 20:45

See, told you Stats is easy.

Hxc · 11 Feb 2010 at 20:47

Do remember that with binomial you will have to have a posh calculator or use the tables if you are working out culumative binomial (ie, what is the chance of 5 or less being cracked) probability, unless you want to spend ages on a normal calc

Stats is however, easy!

Cookeh · 11 Feb 2010 at 20:47

Pudney@work said:
See, told you Stats is easy.

Can I ask you guys another question on it then? xD

sigma · 11 Feb 2010 at 20:54

Combinations and permuations are where it's at - although not in this case heh.

G-Dubs · 11 Feb 2010 at 20:57

G-Dubs said:
Never heard of binomunal maths, so guess this is "probably" wrong, the probability of a cracked egg is 5% which would be easy if there were 20 eggs in a basket (1 egg cracked)

So number of eggs cracked = 1/2 of an egg

That's about as far as I can get yer buddy.

So I was right?

Wow! Now why couldn't I do that 30 years ago?

ordinaryjoe · 11 Feb 2010 at 22:06

I can't think of why simply multiplying 0.05 by 10 would be incorrect but I would think the probability of no cracked eggs would be worked out by doing (1-0.05)^10, as the chance of the first egg being uncracked is 1 - 0.05 which is 0.95, then the probability of the first and second being uncracked would be 0.95^2 etc etc. This would give an answer of 0.599 to 3sf chance of no eggs being cracked. However, I can't think of a logical reason why simply multiplying 0.05 by 10 is wrong...

iraiguana · 11 Feb 2010 at 22:35

The independent portion of the question should rule out the above surely? ie each egg has its own probability and not dependant on a series of 10 eggs being found not cracked?

Or am I miles out here ?

ordinaryjoe · 11 Feb 2010 at 22:45

yes but if any of the previous eggs are cracked, it doesn't matter whether the next eggs are cracked or not (to find the probability of no eggs being cracked). Therefore, we only need to consider the probability of the next egg being cracked when none of the previous eggs are cracked. Think of it like this:

egg 1 cracked?? Yes --->At least 1 egg cracked (5% chance)
No (95% chance)
|
v
egg 2 cracked?? Yes --->At least 1 egg cracked (95% x 5% chance)
No (95% x 95% chance)
|
v
egg 3 cracked?? Yes --->At least 1 egg cracked (95% x 95% x 5% chance)
No (95% x 95% x 95% chance)
|
v
etc etc until egg 10 is not cracked

So the chances are 0.95^10

Vampyre · 11 Feb 2010 at 23:06

G-Dubs said:
Never heard of binomunal maths, so guess this is "probably" wrong, the probability of a cracked egg is 5% which would be easy if there were 20 eggs in a basket (1 egg cracked)

So number of eggs cracked = 1/2 of an egg

That's about as far as I can get yer buddy.

Surely if half an egg is cracked then the whole egg is cracked, so the answer is clearly 0.1

G-Dubs · 13 Feb 2010 at 01:16

Vampyre said:
Surely if half an egg is cracked then the whole egg is cracked, so the answer is clearly 0.1

Just been to tesco and spent £20 on eggs. I've now had a go at over 300 eggs and I a can confirm that you cannot crack half an egg!!!

May have had a few beers also!