The probability that an individual egg in a carton of eggs is cracked is 0.05. You have picked
out a carton of 1 dozen eggs (that’s 12 eggs) at the grocery store. Determine the probability
that at most two of the eggs in the carton are cracked
To determine the probability that at most two of the eggs in the carton are cracked, we can use the binomial probability formula.
The binomial probability formula is:
P(x) = (nCx) * (p^x) * (q^(n-x))
Where:
P(x) is the probability of getting exactly x successes
n is the number of trials or observations (in this case, the number of eggs in the carton)
x is the number of successes (in this case, the number of cracked eggs)
p is the probability of success (in this case, the probability that an individual egg is cracked)
q is the probability of failure (in this case, the probability that an individual egg is not cracked)
Let's calculate the probability that at most two eggs in the carton are cracked using the formula.
To find the probability that exactly 0, 1, or 2 eggs are cracked, we need to calculate the individual probabilities for each case and then add them together.
P(0 cracked eggs) = (12C0) * (0.05^0) * (0.95^(12-0))
P(1 cracked egg) = (12C1) * (0.05^1) * (0.95^(12-1))
P(2 cracked eggs) = (12C2) * (0.05^2) * (0.95^(12-2))
Let's calculate each probability separately.
P(0 cracked eggs) = (12C0) * (0.05^0) * (0.95^12)
The combination (nCx) with x = 0 is equal to 1, so we can omit it.
P(0 cracked eggs) = (0.05^0) * (0.95^12) = 0.95^12 ≈ 0.6826
P(1 cracked egg) = (12C1) * (0.05^1) * (0.95^11)
The combination (nCx) with x = 1 is equal to 12, so we can substitute it.
P(1 cracked egg) = (12) * (0.05^1) * (0.95^11) ≈ 0.2947
P(2 cracked eggs) = (12C2) * (0.05^2) * (0.95^10)
Using the combination (nCx) with x = 2, we find:
P(2 cracked eggs) = (66) * (0.05^2) * (0.95^10) ≈ 0.0718
Finally, to find the probability of at most two cracked eggs (P(at most 2 cracked eggs)), we add these probabilities together:
P(at most 2 cracked eggs) = P(0 cracked eggs) + P(1 cracked egg) + P(2 cracked eggs)
P(at most 2 cracked eggs) ≈ 0.6826 + 0.2947 + 0.0718 ≈ 1.0491
However, probabilities should not exceed 1. Therefore, the actual probability of at most two eggs in the carton being cracked is 1.
So, the probability that at most two of the eggs in the carton are cracked is 1.