Question

PROBABILITY QUESTION

men that would stop and ask for direction : 42%
women that would stop and ask for direction : 61%

Population estimation,
men : 48.2%
women : 51.8%

1) calculate probability driver stops and ask for direction
2) given driver stops to ask for directions,
determine probability that the driver was a man.

Answer 1

Okay so let \(A\) be the event you stop and ask for directions and let \(B\) be the event you're a man.

Answer 2

"men that would stop and ask for direction : 42%"\[
\Pr(A|B)=0.42
\]

Answer 3

"women that would stop and ask for direction : 61%"\[
\Pr(A|B^C)=0.61
\]

Answer 4

okay

Answer 5

Population estimation,
men : 48.2%
women : 51.8% \[
\Pr(B)=0.482\\
\Pr(B^C)=0.518
\]

Remember when someone says \[\Huge
\text{Women and minorities}
\]What they are actually saying is \[\Huge
\text{Everyone}
\]because men are a minority.

Answer 6

Okay so finally, you need to remember that: \[
\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}
\]

Answer 7

In this case, it helps to realize \[
\Pr(A\cap B)=\Pr(A|B)\times \Pr(B)
\]

Answer 8

"1) calculate probability driver stops and ask for direction"
What they want is \(\Pr(A)\).
We need to use total probability here: \[
\Pr(A)= \Pr(A\cap B) + \Pr(A\cap B^C)
\]

Answer 9

1.03?

Answer 10

"2) given driver stops to ask for directions,
determine probability that the driver was a man."
This asks for \(\Pr(B|A)\).
It's going to help to know \(\Pr(A)\) for this one since: \[
\Pr(B|A)=\frac{\Pr(B\cap A)}{\Pr(A)}
\]

Answer 11

but i thought probability cant be more than 1?

Answer 12

You didn't do it right.

Answer 13

Show me your calculation.

Answer 14

0.42 + 0.61

Answer 15

is that wrong? for the first question

Answer 16

or do i have to divide with the total sum?

Answer 17

but if its by % the total sum should be one right?
im quite confuse

Answer 18

Okay what you have is \[
\Pr(A|B)+\Pr(A|B^C)
\]You want to find \[

\Pr(A\cap B)+\Pr(A\cap B^C)
\]

Answer 19

Remember that \[
\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}
\]

Answer 20

You're confusing conditional probability with joint probability.

Answer 21

Do I need to spell it out more?

Answer 22

\[
\Pr(A\cap B) = \Pr(A|B)\times \Pr(B) = (0.42)(0.482)
\]

Answer 23

It's very similar for \(\Pr(A\cap B^C)\)

Answer 24

ah okay i see

Answer 25

thank you so much!
@wio