Question

A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" Find the probability that the number of unspoiled eggs among the selected is a. exactly Round your answer to four decimal places. exactly b. or fewer Round your answer to four decimal places. or fewer c. more than Round your answer to four decimal places. more than

Answer 1

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let X = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

$P(X)=p=\frac{10}{18}=0.556$

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of X is given by:

$P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...$

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

$P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531$

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              $=\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959$

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              $=1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747$

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.