Urgent: Maths CIE Stats

[$H00t!N& $t@r]

sorry guys i have another question.. 

Eggs produced at a farm are packaged in boxes of six. Assume that, for any egg, the probability that it is broken when it reaches the retail outlet is 0.1 , independent of all other eggs. A box is said to be bad if it contains at least two broken eggs. Calculate the probability that a randomly selected box is bad.

Ten boxes are chosen at random. Find the probability that just two of these boxes are bad.

It is known that, in fact, breakages are more likely to occur after the eggs have been packed into boxes, and while they are being transported to the retail outlet. Explain why this fact is likely to invalidate the calculation.

[$H00t!N& $t@r]

the answer is as follows:

0.114; 0.2223   
leave the last part of the question... i understand the answer now

[$H00t!N& $t@r]

Give me 10 minutes, i will explain.

alrite thank you! 

[lana]

sorry guys i have another question..  

Eggs produced at a farm are packaged in boxes of six. Assume that, for any egg, the probability that it is broken when it reaches the retail outlet is 0.1 , independent of all other eggs. A box is said to be bad if it contains at least two broken eggs. Calculate the probability that a randomly selected box is bad.

Ten boxes are chosen at random. Find the probability that just two of these boxes are bad.

It is known that, in fact, breakages are more likely to occur after the eggs have been packed into boxes, and while they are being transported to the retail outlet. Explain why this fact is likely to invalidate the calculation.

(i) at least two eggs broken: so we find the probability that less than two eggs and then subtract it from one :
(6C1) (0.1)^1 (0.9)^5 + (6C0)(0.1)^0 (0.9)^6 = 0.8857
1- 0.8857=0.144

(ii) from the ten boxes they want the probability of two boxes :
(10C2) ( 0.144)^2 (0.8857)^8 = 0.223

(iii) the event that eggs are broken is not independent of the event that any other egg is broken ( not sure )

i hope it's right because I'm not really sure !!! hope it helps =]

[Freaked12]

+rep

[$H00t!N& $t@r]

(i) at least two eggs broken: so we find the probability that less than two eggs and then subtract it from one :
(6C1) (0.1)^1 (0.9)^5 + (6C0)(0.1)^0 (0.9)^6 = 0.8857
1- 0.8857=0.144

(ii) from the ten boxes they want the probability of two boxes :
(10C2) ( 0.144)^2 (0.8857)^8 = 0.223

(iii) the event that eggs are broken is not independent of the event that any other egg is broken ( not sure )

i hope it's right because I'm not really sure !!! hope it helps =]

Thanks a lot again!! +rep again 

[lana]

your sooo welcome I'm just happy to help  
thank youu guys for +rep  

[Deadly_king]

Explained it in a easier way

~Sf staff at your service

I like that

[TJ-56]

Can some1 help!?
may/june 2010 paper 61 q5 (ii)
Please explain, the mark scheme is useless on this one
thanx in advance

[elemis]

Can some1 help!?
may/june 2010 paper 61 q5 (ii)
Please explain, the mark scheme is useless on this one
thanx in advance

Read the attached Examiners Report for help.

[lana]

soo how did you guys do ??  I sat for 62 and I thought it was very tricky 

[Deadly_king]

soo how did you guys do ??  I sat for 62 and I thought it was very tricky 

Yeah.......i sat for the same paper as well.

I made some monstrous mistakes that I can't help imagine how i managed to do that

It sure was tricky

[lana]

Yeah.......i sat for the same paper as well.

I made some monstrous mistakes that I can't help imagine how i managed to do that

It sure was tricky

I DID THE SAME THING  
can you believe that anyone could get the approximation question wrong i  did !!! i forgot to put .5 thing  
what did you get for the second question the standard deviation, mean and the ?(x-50) and
?(x-50)^2 ??

[Deadly_king]

I DID THE SAME THING  
can you believe that anyone could get the approximation question wrong i  did !!! i forgot to put .5 thing  
what did you get for the second question the standard deviation, mean and the ?(x-50) and
?(x-50)^2 ??

Am not sure about the approximation too

I can't remember the mean but I think i got the summation to be 712.

Tell me your values, i'll see if i remember

[lana]

for the summation of (x-50) i got 24 and for the summation of (x-50)^2 i got the same thing 712 or 715 something like that but i cant remember the exact number !! for the s.d i got 6. something or 7.something and for the new mean i got 52.7 !!
and how about the last question (iii) ?? i got 9030 and I'm pretty sure it's wrong !!