ANSWER:
$1.8
Step-by-step explanation:
Given:
A bag of 10 fortune cookies;
Five fortune cookies contain "$1 off"
Four contain "$2 off"
One contains "$5 off"
So the probability of selecting a "$1 off" cookie will be;
![\begin{gathered} P(\text{\$1 }off\text{ }cookie)=(5)/(10) \\ =(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4ckez35k0vfcxl8qbpxf8ahqztn5e9ll7x.png)
And the probability of selecting a "$2 off" cookie will be;
![\begin{gathered} P(\text{\$2 }off\text{ }cookie)=(4)/(10) \\ =(2)/(5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/z3gh4s3w64w5j72akpa3500rjc9ei5fskx.png)
the probability of selecting a "$5 off" cookie will be;
![\begin{gathered} P(\text{\$5 }off\text{ }cookie)=(1)/(10) \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f96stba890jjoiihfkc42eu5dh96m1eafe.png)
We can determine the expectation of a selection as seen below;
![\begin{gathered} Expectation=(1*(1)/(2))+(2*(2)/(5))+(5*(1)/(10)) \\ \\ =(1)/(2)+(4)/(5)+(1)/(2) \\ \\ =(9)/(5) \\ \\ =\text{ \$}1.8\text{ }off \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ms7zynms720u79s92f1sobxd2oizo0t9ne.png)
So the expectation of a selection will be $1.8 off