91影视

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:

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

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

Determine the standard deviation as follows:

$蟽=\sqrt{鈭�y^2脳P\left(y\right)-{\left(鈭�y脳P\left(y\right)\right)}^2}$

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

$=0.822$

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:

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

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:

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

As a result, the probability is$0.295$