T6.11. 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 :	0	1	2	3	4
Probability :	0.78	0.11	0.7	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 mean.
(c) Calculate standard deviation. Show your work.
(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.

Let, there are at least 10 unbroken eggs, only two are broken.
The probability is computed as follows:
$$\begin{gathered} P(Y=2)=P(Y=0)+P(Y=1)+P(Y=2) \\ =0.78+0.11+0.07 \\ =0,96 \end{gathered}$$

As a result, the probability is 0.96

The mean can be calculated as:

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

This indicates that in a dozen eggs, there are on average 0.38 worth of broken eggs.

Determine the standard deviation as follows:

$\sigma =\sqrt{\sum y^2\times P\left(y\right)-{\left(\sum y\times P\left(y\right)\right)}^2}$

$=\sqrt{0^2\times 0.78+1^2\times 0.11+\dots .+4^2\times 0.01-(0.38{)}^2}$

$=0.822$

The number of broken eggs will be 0.822 more than the mean of 0.38 eggs.

Determine the probability of two broken eggs is calculated as follows:

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

Then,

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

Hence, the probability is calculated as:

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

As a result, the probability is$0.295$