Looking for some help on this question. I can't figure out if this is a unconditional or conditional probability and which formulas to use.

Suppose that 18% of the employees of a given corporation engage in physical exercise activities during the lunch hour. Assume that 57% of all employees are male, and 12% of all employees are males who engage in physical exercise activities during lunch hour.

a) If we choose an employee at random from this corporation, what is the probability that this person is a female who engages in physical exercise activities during the lunch hour.

b) If we choose an employee at random from this corporation, what is the probability that this person is a female who does not engage in physical exercise activity during the lunch hour.

I am having difficulties determining if the 12% of employees who are males who engage in physical exercise activities during lunch hour is P(AIB) or if it is P(A and B) and how to answer part a and b.

Any help would be great!

Thanks

a) Not conditional

b) Not conditional

Notice that the questions says 'an employee', meaning that it could be a male or a female chosen, and hence, it doesn't mean 'what is the probability that an employee engages in physical exercise activities during the lunch hour, given the employee is female' or something like that.

If you build a table, you can answer any kind of "what is the probability of something given something" type problem.

Assume an arbitrary number of people. Something easy to work with like 100.

\begin{array}{c|c|c|c} \;\ &\text{male}&\text{female}&\text{total}\\ \hline \text{activity}&12&6&18\\ \hline \text{no activity}&45&37&82\\ \hline \text{total}& 57&43&100\end{array}

Now, suppose it said, "what is the probability an employee chosen was male given they do not engage in activity"?

Go down the 'male' column to 'no activity' row, then go across to the total in that row. \frac{45}{82}

"What is the probability an employee chosen engages in activity given they're female"?

Go across the 'activity' row to the 'female' column and go down.

\frac{6}{43}

And so on...

Thanks so much for the help!

Sorry I am confuse , does not it mean that 12 % of ( all emp who are engages in physical activity ) are male which means 12 % of 18 % are male

No, not 12% but 12 of 18 employees who engage in physical activity are male.

The percentage is (12/18)*100% = 66.7%

I am confused. How does this fit in the probability formula - P(A and B) = P (A|B) P (B)?

NO. It states 12% of all employees are males who engage in physical exercise activities during lunch hour. Does this mean 12 of 18 employees? I am confused.

I am confused. How does this fit in the probability formula - P(A and B) = P (A|B) P (B) ?.

There is no need to get to conditional probability in this question!! galactus only showed how to answer the question if ever it required conditional probability.

When he said

"what is the probability an employee chosen was male given they do not engage in activity"?

You are looking for P(M\ |\ \bar{A})

So, this is given by: P(M\ |\ \bar{A}) = \frac{P(M\ \cap\ \bar{A})}{P(\bar{A})}

P(M\ \cap\ \bar{A}) = 45%

P(\bar{A}) = 82%



NO. It states 12% of all employees are males who engage in physical exercise activities during lunch hour. Does this mean 12 of 18 employees?. I am confused.

No, read the given again.

"12% of all employees are males who engage in physical exercise activities during lunch hour."

All employers... this means all of the 100% employees. Or 12% of the TOTAL number of employees. Note that we don't know the exact number of employees doing such and such thing.

But in an earlier posting you mentioned this -
No, not 12% but 12 of 18 employees who engage in physical activity are male.

The percentage is (12/18)*100% = 66.7%

Yes, and what does 12 of 18 employees mean? It means that for every 18 employees, there are 12 who engage in physical activity. That's what the data told us.

OK.. so what will be the correct calculation for this?
b) If we choose an employee at random from this corporation, what is the probability that this person is a female who does not engage in physical exercise activity during the lunch hour.

Is it 37/43 OR 37/82 ?

b) If we choose an employee at random from this corporation, what is the probability that this person is a female who does not engage in physical exercise activity during the lunch hour.

I underlined/bolded the key terms. "this corporation" means the whole lot of employees, the 100%.

This means that the probability is 37/100. Hence why I told you that there was no conditional probability involved from the very start.

Was that so hard?

Sure, but I was confused with Galactus's calculations above. If what you say is correct, then the answer to the other question should be 6/100?
a) If we choose an employee at random from this corporation, what is the probability that this person is a female who engages in physical exercise activities during the lunch hour.

Do you still need help with this?

Sure, but I was confused with Galactus's calculations above. If what you say is correct, then the answer to the other question should be 6/100?
a) If we choose an employee at random from this corporation, what is the probability that this person is a female who engages in physical exercise activities during the lunch hour.

Right, you got the correct answer.

galactus only told you what to do if ever the question was a conditional probability question.

Thanks for your help. I really appreciate.

No.. I am all set. Thanks.