Hello,

I have one problem with distinguishing the conditional probability from the intersection.
As an example I uploaded one problem in the jpg file.
I solved the problem, but it took me quite a while to figure it out.
My main problem here was:
The sentence: Only 30% of unsuccessful employees had passed.

How do I know that this sentence doesn't tell me that

The Probability of the intersection of unsuccessful employees and employees who passed the test is 30%

and that actually it means that

The Probability that an employee has passed, given that he or she is unsuccessful is 30%

 

VVS,

Am I overlooking something here? Doesn't it say that 65% of the recruits will be successful, thereby answering the last sentence of the problem? One has to assume that no one will be hired unless they pass the test. So isn't the probability 0.65?

Ratch

 

Looking at the question logically, if unsuccessful employees are defined as those who have not passed(=failed), then the statement that 30% of them have passed is a non sequitor (or something!) There is an element missing in the statement that is needed for it to make any logical sense. The word "previously" before unsuccessful would make sense, otherwise I don't see how this statement tells you anything!
Colin1928

 

If you were to say this question was poorly worded, I would agree.

To solve it you need to realise the universal set or population set is that of recruits.
We are not told that all passing an aptitude test are employed or that all failing are not employed, only a percentage in each case.

To make the numbers easy I have taken a group of 1000 recruits. We are given that 650 of these will be employed (=successful) and 350 not employed (=unsuccessful).

The wording '....successful employees, and of these 80%...' tells us that 80% of 650 (=520) has passed the test and 130 had failed.

Equally the wording '.....30% of unsuccessful...' tells us that 30% of 350 (=105) had passed the test and 245 had failed.

Now we are asked for the probability of gaining employment within the ranks of those who had passed the test. This is the same as the % of those employed related to the total passes i.e 520 /(520+105), when divided by 100 to make a fraction.

Here I again want to create a table, but am limited by the forum format. However I have tabulated the possibilities

........................Pass........Fail.......Totals

Employed...........520..........130.......650

Not Employed.....105...........245......350

Totals...............625...........375.....1000

The conditional probability we are looking for is 520/625 = (0.65*0.80) / .625 = .832

If you want to do it by a probability tree, or by formula you will arrive at the middle expression in my equations above.
My table is equivalent to a Venn Diagram of the situation.

 

Ahaaa! i get it now. thanx a lot everybody!

 

Nicely explained "Studiot"

I was wondering how to draw a diagram to visualize the problem and came up with the attachment.

Interesting that the test was a useful predictor of "success". Fortunately the wise heads in the (hypothetical) recruitment section "took a chance" on some people who failed the test but were still successful. Somewhat like the modern approach to recruiting medical students - people skills like bedside manner & empathy might turn out useful in the real world - not just being in the top 2% of academic achievers.

Ever see the film "Gattaca"?