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  #1
  it has been found that on average 10% of the eggs supplied to a supper market are c racked.if you buy a box of 8eggs what is the probability that it contains 2or More cracked eggs
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  9 months ago
  #2
  X ~ b(8,0.1)
  p(x>=2) = 1 - p(x<=1)
  = 1 - ( 8c0(0.1)^0(0.9)^8 + 8c1(0.1)^1(0.9)^7)
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  #3
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  9 months ago
  #4
  The first few words indicate that you can model this situation binomally.
  A binomial question should satisfy a few conditions:
  - Fixed number of trials (testing 8 eggs in this case)
  - Only two outcomes, success or failure.
  
  When the question said "on average 10% of the eggs supplied to a supper market are cracked", it was another way of saying the probability that a randomly selected egg is cracked is 10% -> so in this case the success is a cracked egg (probability = 0.1) and the failure is not choosing a cracked egg (probability = 0.9 as probability of success + probability of failure should add up to 1)
  
  Now that we have constructed our Binomial distribution: X~B(n,p) where n is fixed number of trials (8) and p is probability of success (0.1), we can think about the probability of choosing 2 or more eggs, which is 2,3,4,5,6,7 or 8 eggs, however working out each probability and adding them all together is long so we will use the fact that 0,1,2,3,4,5,6,7 or 8 eggs being cracked should add up to one and we reverse the situation by saying P(X>=2) is the same as 1 - P(X<=1).
  
  So, how do you work out P(X<=1)? Well we have our Binomial distribution: X~B(8,0.1) so we just do P(X=0) which is 8C0(0.1)^0(0.9)^8 by using binomial formula of nCr(p)^r(1-p)^n-r where r is number of successes and also P(X=1) which is 8C1(0.1)^1(0.9)^7 by using the same formula then add them together and thats P(X<=1).
  (Original post by Francis Kabwe)
  1
  #5
  Thanks for that dear
  (Original post by ah18212)
  The first few words indicate that you can model this situation binomally.
  A binomial question should satisfy a few conditions:
  - Fixed number of trials (testing 8 eggs in this case)
  - Only two outcomes, success or failure.
  
  When the question said "on average 10% of the eggs supplied to a supper market are cracked", it was another way of saying the probability that a randomly selected egg is cracked is 10% -> so in this case the success is a cracked egg (probability = 0.1) and the failure is not choosing a cracked egg (probability = 0.9 as probability of success + probability of failure should add up to 1)
  
  Now that we have constructed our Binomial distribution: X~B(n,p) where n is fixed number of trials (8) and p is probability of success (0.1), we can think about the probability of choosing 2 or more eggs, which is 2,3,4,5,6,7 or 8 eggs, however working out each probability and adding them all together is long so we will use the fact that 0,1,2,3,4,5,6,7 or 8 eggs being cracked should add up to one and we reverse the situation by saying P(X>=2) is the same as 1 - P(X<=1).
  
  So, how do you work out P(X<=1)? Well we have our Binomial distribution: X~B(8,0.1) so we just do P(X=0) which is 8C0(0.1)^0(0.9)^8 by using binomial formula of nCr(p)^r(1-p)^n-r where r is number of successes and also P(X=1) which is 8C1(0.1)^1(0.9)^7 by using the same formula then add them together and thats P(X<=1).
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