A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F

(a) The likelihood that all 5 eggs chosen are unspoiled is 0.0531. (b) The probability that 2 or fewer out of the 5 eggs are unspoiled is 0.3959. (c) The probability that more than 1 of the selected 5 eggs are unspoiled is 0.8747. Step-by-step explanation: The complete query is: A subpar carton of 18 eggs has 8 that are spoiled. An unsuspecting chef selects 5 eggs at random for his “Mega-Omelet Surprise.” Calculate the probability of receiving (a) exactly 5 unspoiled eggs, (b) 2 or fewer, and (c) more than 1 unspoiled egg. Define X = number of unspoiled eggs. In the faulty carton, 8 eggs are spoiled. The probability of selecting an unspoiled egg is independent of others. Provided that a chef randomly picks 5 eggs, the variable X follows a Binomial distribution with parameters n = 5 and p = 0.556. Success is defined as selecting an unspoiled egg. The probability mass function of X is as follows: (a) Calculate the probability of selecting all unspoiled eggs. Thus, this probability is found to be 0.0531. (b) For 2 or less unspoiled eggs, the probability is computed: P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2), resulting in a probability of 0.3959. (c) For more than 1 unspoiled egg: P (X > 1) = 1 - P (X ≤ 1), yields a final probability of 0.8747.