Question

Approximately 3% of the eggs in a grocery store are cracked If you buy two dozen eggswithout checking them first, what is the probability that at least one of your eggs iscracked ?

Answer 1

This is a binomial probabilty problem

Let X be the event that your eggs are cracked. P(x) = 0.03

P(eggs that are not cracked)=1 - 0.03 = 0.97

For each egg, there is a probability that an egg cracked p = 0.03 that the egg is broken and a probability q = 1 - p = 0.97 that the egg is not cracked.

P(no more than two of your eggs are cracked)= P(zero eggs are cracked)+ P(1 egg is cracked) +P( 2 eggs are cracked)

P(at least one of your eggs are cracked) = P(X=0)

p = 0.03

q = 0.97

`\begin{gathered} nC_rp^rq^(n-r) \\ P(x=0)=24C_00.03^00.97^(24) \\ P(x=0)=0.481 \end{gathered}`

P(at least one of your eggs are cracked) = 1 - 0.481

P(at least one of your eggs are cracked) = 0.519