##### The Binomial Distribution

 Statistics Tutor: None Selected Time limit: 1 Day

Sep 29th, 2015

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

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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Sep 29th, 2015
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Sep 29th, 2015
Dec 8th, 2016
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