Let Y denote the number of broken eggs in a randomly selected carton of one dozen “store brand” eggs at a local supermarket. Suppose that the probability distribution of Y is as follows.

Value$\left(y_i\right)$	0	1	2	3	4
Probability$\left(P_i\right)$	$0.78$	$0.11$	$0.07$	$0.03$	$0.01$

a. What is the probability that at least 10 eggs in a randomly selected carton are unbroken?

b. Calculate and interpret ${\mu }_Y$.

C. Calculate and interpret ${\sigma }_Y$.

d. A quality control inspector at the store keeps looking at randomly selected cartons of eggs until he finds one with at least 2 broken eggs. Find the probability that this happens in one of the first three cartons he inspects.

The following is the probability distribution:

At most two eggs are broken if at least ten eggs are undamaged.

The following formula can be used to compute the probability:

$\begin{array}{rcl}P(Y& =& 2)=P(Y=0)+P(Y=1)+P(Y=2)\\ & =& 0.78+0.11+0.07\\ & =& 0.96\end{array}$

Thus, the required probability is $0.96.$

The following is the probability distribution:

The average can be calculated as follows:

$\begin{array}{rcl}\text{Mean}& =& \sum y\times P\left(y\right)\\ & =& 0(0.78)+1(0.11)+2(0.07)+3(0.03)+4(0.01)\\ & =& 0.38\end{array}$

The predicted number of broken eggs are 0.38

The following is the probability distribution:

Calculate the standard deviation value,

The number of broken eggs is expected to differ by $0.822$ from the mean of $0.38$ eggs.

The following is the probability distribution:

The chances of receiving at least two cracked eggs are:

$\begin{array}{rcl}P(Y& \ge & 2)=P(Y=2)+P(Y=3)+P(Y=4)\\ & =& 0.07+0.03+0.01\\ & =& 0.11\end{array}$

Now,

$$\begin{gathered} P(X=1)=0.11(1-0.11{)}^{1-1}=0.11 \\ P(X=2)=0.11(1-0.11{)}^{2-1}=0.0979 \\ P(X=3)=0.11(1-0.11{)}^{3-1}=0.087131 \end{gathered}$$

Thus, the resultant probability is:

$\begin{array}{rcl}P(1& \le & X\le 3)=P(X=1)+P(X=2)+P(X=3)\\ & =& 0.11+0.0979+0.087\\ & =& 0.295\end{array}$