Question

Approximately 3% of the eggs in a store are cracked. If you buy 1 dozen eggs, what is the probability that

a) none of your eggs are cracked

b) at least one of your eggs is cracked

c) exactly two of your eggs are cracked

Answer 1

Solution

Given that ,

p = 0.03

1 - p = 0.97

n = 12

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

a)

P(X = 0) = ((12! / 0! (12-0)!) * 0.03⁰ * (0.97)^12-0

=  ((12! / 0! (12)!) * 0.03⁰ * (0.97)¹²

= 0.6938

Probability = 0.6938

b)

P(X 1) = 1 - P( X < 1 )

= 1 - P( X = 0)

= 1 - ((12! / 0! (12-0)!) * 0.03⁰ * (0.97)^12-0

= 1 - ((12! / 0! (12)!) * 0.03⁰ * (0.97)¹²

= 1 - 0.6938

= 0.3032

Probability = 0.3032

c)

P(X = 2) = ((12! / 2! (12-2)!) * 0.03² * (0.97)^12-2

=  ((12! / 2! (12)!) * 0.03² * (0.97)¹⁰

= 0.0438

Probability = 0.0438