The Binomial Distribution

Statistics
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Sep 29th, 2015

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18 a .P(one egg is not cracked or broken)= 1- 0.025 = 0.975

  P(no eggs are not cracked or broken) = (0.975)^12

                                                                      =0.738 (3dp)

18 (b). P(at least egg is cracked or broken) = 1- P(all dozen eggs are not cracked or broken)

                                                              =1 -0.738

                                                               = 0.262

18(b) (ii) P (one egg is cracked or broken) =  12C1 (0.975)^11 x (0.025)

                                                                    = 0.227 (3dp)

 P(two eggs are cracked or broken) = 12C2 (0.975)^10 x (0.025)²

                                                     = 12 x 11/2! x (0.975)^10 x (0.025)²

                                                      = 66 x (0.975)^10 x (0.025)²

                                                      = 0.032

Hence using 18(a) and 18(b) part (I)

P(no eggs or exactly one egg is cracked or broken) = 0.738 + 0.227

                                                                                  = 0.965

Hence P(at least two will be cracked or broken) = 1 -0.965

                                                                            = 0.035

18 (c) Average number of broken eggs = 12 x P(one egg is broken)

                                                    = 12 x .025

                                                        0.3


Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 29th, 2015

Easier way to do 18(b) (ii)

P(at least two broken or cracked) = P(at least one  broken or cracked) - P(exactly one cracked)

                                                     = 0.262 - 0.227 

                                                  =0.035

Sep 29th, 2015

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