# An egg distributer determines that the probability that any individual egg has a

An egg distributer determines that the probability that any individual egg has a crack is 0.15.
​a) Write the binomial probability formula to determine the probability that exactly x eggs of n eggs are cracked.
​b) Write the binomial probability formula to determine the probability that exactly 2 eggs in a​ one-dozen egg carton are cracked. Do not evaluate.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

cheekabooy
Step 1
Some characteristics of binomial distribution:
Each trial has exactly two possible outcomes.
There are n number of trials, in which p is the probability of success and $q\left(=1-p\right)$ is the probability of failure.
The trials are independent of each other.
The probability of success in each trial remains constant.
Formula for binomial probability:
The formula for binomial probability is as follows.
$P\left(X=x\right){=}^{n}{c}_{x}{p}^{x}{q}^{\left(n-x\right)}$
where,
p is the probability of success,
q is the probability of failure,
n is the number of questions
Step 2
a) Binomial probability formula to determine the probability that exactly x eggs of n eggs are cracked:
Here, the event of success is “eggs that have cracks”.
The probability of success (an individual egg has a crack) is, $p=0.15$.
The probability of failure is, $q=1-0.15=0.85$.
The binomial probability formula to determine the probability that exactly x eggs of n eggs are cracked is as follows.
$P\left(X=x\right){=}^{n}{c}_{x}{\left(0.15\right)}^{x}{\left(0.85\right)}^{\left(n-x\right)}$
Step 3
b) Binomial probability formula to determine the probability that exactly 2 eggs in a one-dozen egg carton are cracked:
In general, number of eggs in one-dozen egg carton is, $n=12$.
The binomial probability formula to determine the probability that exactly 2 eggs in a one-dozen egg carton are cracked is as follows.
$P\left(X=2\right){=}^{12}{C}_{2}{\left(0.15\right)}^{2}{\left(0.85\right)}^{\left(12-2\right)}$
${=}^{12}{C}_{2}{\left(0.15\right)}^{2}{\left(0.85\right)}^{10}$