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 distribution.
The binomial distribution is used when there are two possible outcomes for each trial, and the probability of success (cracked egg) remains the same for each trial.
In this case, the probability of success (p) is 0.05, and the number of trials (n) is 12 (the number of eggs in the carton).
To find the probability of at most two eggs being cracked, we need to calculate the probability of 0, 1, or 2 cracked eggs.
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)
To calculate the probability for each case, we can use the binomial probability formula for X successes in n trials:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
where C(n, k) is the binomial coefficient given by C(n, k) = n! / (k!(n-k)!)
Let's calculate the probability for each case:
P(X = 0) = C(12, 0) * (0.05)^0 * (1-0.05)^(12-0)
P(X = 1) = C(12, 1) * (0.05)^1 * (1-0.05)^(12-1)
P(X = 2) = C(12, 2) * (0.05)^2 * (1-0.05)^(12-2)
Now, we need to calculate the binomial coefficient for each case:
C(12, 0) = 12! / (0!(12-0)!) = 1
C(12, 1) = 12! / (1!(12-1)!) = 12
C(12, 2) = 12! / (2!(12-2)!) = 66
Let's substitute the values into the formula and calculate the probabilities:
P(X = 0) = 1 * (0.05)^0 * (1-0.05)^(12-0)
P(X = 1) = 12 * (0.05)^1 * (1-0.05)^(12-1)
P(X = 2) = 66 * (0.05)^2 * (1-0.05)^(12-2)
Now, sum up the probabilities to find P(X ≤ 2):
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
Finally, calculate the probability:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)