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
Thanks for you help! But I am still SO Lost!
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For a) you said the probability of no cracked eggs is .
.........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?
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For b) you said
........where p= probability of success and n=?
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.