Probability of getting a cracked egg at the store is 7%

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February 10th 2013, 08:57 AM
tdotodot

Probability of getting a cracked egg at the store is 7%

Probability of getting a cracked egg at the store is 7%:

a) What is the probability that a caron of 12 eggs will contain at LEAST one cracked egg? (hint: use indirect method to answer).

b) Determine the number of cracked eggs you expect to find in a gross of eggs? (A 'gross' is 12 dozen OR 144).

I am totally LOST :(. Would I be using Binomial Distribution with this formula: p(x)=C(n,x) · p^x · q^n-x
February 10th 2013, 09:16 AM
Plato

Re: Probability of getting a cracked egg at the store is 7%

Quote:

Originally Posted by tdotodot

Probability of getting a cracked egg at the store is 7%:
a) What is the probability that a caron of 12 eggs will contain at LEAST one cracked egg? (hint: use indirect method to answer).

b) Determine the number of cracked eggs you expect to find in a gross of eggs? (A 'gross' is 12 dozen OR 144).

a) The probability of no cracked eggs in 12 is $(0.93)^{12}$ .

b) For binomial probability the expected value is $E=p\cdot N$ .
February 11th 2013, 09:36 AM
tdotodot

Re: Probability of getting a cracked egg at the store is 7%

Quote:

Originally Posted by Plato

a) The probability of no cracked eggs in 12 is $(0.93)^{12}$ .

b) For binomial probability the expected value is $E=p\cdot N$ .

Thanks for you help! But I am still SO Lost!
__
For a) you said the probability of no cracked eggs is $(0.93)^{12}$ .
.........would I subtract that by 1 to get the probability of having at least one cracked egg?
.........how do I subtract that thing(decimal to a power) by 1?
.........how do I express a decimal to the power of something in a percentage?

__
For b) you said $E=p\cdot N$
........where p= probability of success and n=?
February 11th 2013, 09:46 AM
Plato

Re: Probability of getting a cracked egg at the store is 7%

Quote:

Originally Posted by tdotodot

For a) you said the probability of no cracked eggs is $(0.93)^{12}$ .
.........would I subtract that by 1 to get the probability of having at least one cracked egg?
.........how do I subtract that thing(decimal to a power) by 1?
.........how do I express a decimal to the power of something in a percentage?
For b) you said $E=p\cdot N$
........where p= probability of success and n=?

Are you serious? Are these really questions you have?
I mean, surely you can do grade school arithmetic if you are working probability problems.

a) $1-(0.93)^{12}$ . Use a calculator.

b) $144(0.07)$ . Use a calculator.
February 11th 2013, 11:19 AM
tdotodot

Re: Probability of getting a cracked egg at the store is 7%

Quote:

Originally Posted by Plato

Are you serious? Are these really questions you have?I mean, surely you can do grade school arithmetic if you are working probability problems.a) $1-(0.93)^{12}$ . Use a calculator.b) $144(0.07)$ . Use a calculator.

I don't know, I think it's the fact that I'm not feeling well, plus I've been working on this question plus another questions I'm stuck on for over an hour with no luck (before you came by).I still have no idea how you got (0.93)^12I'm sorry If I sound stupid. :(Edit: OH WAIT! 0.07 (is the 7 percent of one cracked egg.) So 1 - 0.07 = 0.93. RIGHT? But then you make it to the power of 12, becasue there 12 eggs in a carton? ......See I'm not quite following the theory behind this. In other words you guided me through this question but if I saw something similar I don't think I'd get very far.
February 11th 2013, 11:36 AM
tdotodot

Re: Probability of getting a cracked egg at the store is 7%

Alright I finally figured it out, and understand it.

THANKS FOR YOUR HELP PLATO! :D