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 <em>X</em> = 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:
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 <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.
The success is defined as the selection of an unspoiled egg.
The probability mass function of <em>X</em> is given by:
(a)
Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:
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)
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)
Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.
