Problem Solving:Using the Binomial Formula: P(x)=(nCx)pxqn-x
A Dozen Eggs An egg distributor determines that the probability that any individual egg has a crack is .014.
a)Write the binomial probability formula to determine the probability that exactly x of n eggs are cracked.
b)Write the binomial probability formula to determine the probability that exactly 2 in a one-dozen egg carton are cracked.
px=.014 qx=.986 nCx= n!/x!(n-x)!
What is your question? I will be happy to critique your thinking or work.
a) The binomial probability formula to determine the probability that exactly x of n eggs are cracked is P(x) = (nCx) * p^x * q^(n-x), where n is the total number of trials (in this case, the number of eggs), x is the number of successes (cracked eggs), p is the probability of success (the probability that any individual egg has a crack), q is the probability of failure (1 - p), and (nCx) represents the number of ways to choose x items from a set of n items.
b) In this case, we are considering a one-dozen egg carton, which means n = 12 (the total number of eggs). We want to find the probability that exactly 2 eggs are cracked, so x = 2.
Using the given information:
p = 0.014 (the probability that any individual egg has a crack)
q = 1 - p = 1 - 0.014 = 0.986
nCx = n!/x!(n-x)! = 12! / 2!(12-2)! = 12! / 2!10! = (12 * 11) / 2 = 66
Plugging these values into the formula, we have:
P(x) = (nCx) * p^x * q^(n-x) = 66 * (0.014)^2 * (0.986)^(12-2) = 66 * 0.000196 * 0.823
To find the final probability, simply calculate the result:
P(x) = 0.000196 * 0.823 = 0.000161308 (approximately)
Therefore, the probability that exactly 2 eggs in a one-dozen egg carton are cracked is approximately 0.000161308.