You question is of two parts.
For the first part:
A group of men and women were asked what their favorite pet was, and the results of the survey were tabulated.
[tex]\begin{center}
\begin{tabular}
{|c|c|c|c|c|}
& Cats & Dogs & Other & Total \\ [1ex]
Male & 42 & 58 & 6 & 106 \\
Female & 52 & 48 & 2 & 102 \\ [1ex]
Total & 94 & 106 & 8 & 208
\end{tabular}
\end{center}[/tex]
Let event A be defined as randomly
choosing someone who picked cats or dogs as their favorite pet. Let
event B be defined as a randomly
chosen person being male.
Find P(B| NOT A).
P(A) = P(choosing cat) + P(choosing dog) = [tex] \frac{94}{208} + \frac{106}{208} = \frac{200}{208} = \frac{25}{26}[/tex]
P(NOT A) = [tex]1- \frac{25}{26} = \frac{1}{26} [/tex]
P(B and NOT A) = P(males that did not choose cat or dog) = [tex] \frac{6}{208} = \frac{3}{104} [/tex]
P(B | NOT A) = [tex] \frac{P(B \, and \, NOT \, A)}{P(NOT \, A)} = \frac{ \frac{3}{104} }{ \frac{1}{26} } =\frac{3}{104}\times26= \frac{3}{4} [/tex]
For the second part of the question.
Petey
is considering investing $19 in a certain company. Financial advisors
forecast that there is a 30% chance that the stock will increase in
value by 10%, and a 70% chance he will lose his initial investment.
Determine if Petey should make the investment, and find the expected
value of the investment.
If his investment increases by 10%, the value of the investment will be 1.1 x $19 = $20.90 with a probability of 30% or 0.3
The expected value of the investment is given by
[tex]\$20.90\times0.3+(-\$19\times0.7)=\$6.27-\$13.30=-\$7.03[/tex]
Therefore, Petey should not make the investment as there is an expectation of a loss from the investment.
