Respuesta :
Answer:
The probability that at most 3 eggs are cracked is 0.9689.
Step-by-step explanation:
The probability of an event E is defined as:
[tex]P(E)=\frac{Favorable\ outcomes}{Total\ no.\ of\ outcomes}[/tex]
A gross of eggs consists of N = 144 eggs.
Number of cracked eggs (X) in the gross is, k = 12 eggs.
Compute the probability of selecting a cracked egg as follows:
[tex]P(X)=\frac{k}{N}=\frac{12}{144}=0.0833[/tex]
A random sample of n = 15 eggs are selected for inspection.
Then the random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of a Binomial distribution is:
[tex]P(X=x)={15\choose x}0.0833^{x}(1-0.0833)^{15-x};\ x=0,1,2,3...[/tex]
Compute the probability that at most 3 eggs are cracked as follows:
P (X ≤ 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
[tex]={15\choose 0}0.0833^{0}(1-0.0833)^{15-0}+{15\choose 1}0.0833^{1}(1-0.0833)^{15-1}\\+{15\choose 2}0.0833^{2}(1-0.0833)^{15-2}+{15\choose 3}0.0833^{3}(1-0.0833)^{15-3}\\=0.2713+0.3698+0.2352+0.0926\\=0.9689[/tex]
Thus, the probability that at most 3 eggs are cracked is 0.9689.
