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Independent Outcomes OR (AND?) Dependent Ones

Date: 08/21/2012 at 04:21:39
From: Millicent 
Subject: Do mutually exclusive events always have to be an OR

I was given this problem:

At a certain non-profit organization, 56% of employees are college
graduates and 53% of employees have more than ten years of experience. If
75% of the organization's employees are either college graduates or have
more than ten years of experience (or both), what is the probability that
a randomly selected employee will have more than ten years of experience
and be a college graduate?

I am confused about which formula I am supposed to use for this problem:

   P(A)*P(B)      or      P(A) + P(B) - P(A and B)? 

And why?

I know that the probability of event A AND event B occurring is P(A)*P(B).
So I tried doing this:

   P(employee will have more than ten years of experience 
     AND be a college graduate)
     = 0.56*0.53

But the answer told me that I had to use the probability addition formula,
and that I have to add the probability of the two events, then subtract
the probability that both events occur: (0.56*0.53 - 0.75). I thought that
that formula is only used when the problem says OR, not AND. (For example,
what is the probability that you get an 8 OR a hearts from a deck of cards?
P = (4/52) + (13/52) - (1/52).)

Then I thought that maybe the formula P(A and B)=P(A)*P(B) only works when
there are two separate events, such as roll a 6, and then roll a 5. In the
original problem, there is only one event -- namely, whether the ONE
employee is a college graduate and has more than 10 years of experience.
If this problem were to ask if you pick two college employees, and what is
the probability that the first is a college graduate and the second has
more than 10 years of experience, then would you use the formula I
mentioned above?

Please explain in detail when each formula is supposed to be used. I am
extremely confused about why this problem asks you to use the probability
addition formula.

Sorry for the really confusing explanation. Thank you so much for your help.



Date: 08/21/2012 at 10:48:34
From: Doctor Peterson
Subject: Re: Do mutually exclusive events always have to be an OR

Hi, Millicent.

Just remember the entire equation, not one side only -- and don't forget
the conditions required:

   If A and B are ANY events, then
   P(A and B) = P(A) * P(B | A)      [you may not have seen this yet]
 
   If A and B are INDEPENDENT events, then
   P(A and B) = P(A) * P(B)             [the convenient special case]
 
   If A and B are ANY events, then
   P(A or B) = P(A) + P(B) - P(A and B)            [the general case]
 
   If A and B are MUTUALLY EXCLUSIVE events, then
   P(A or B) = P(A) + P(B)    [the special case where P(A and B) = 0]
 
These are not just things to do when events are independent or mutually
exclusive; they are ways to find the "or" or the "and" when that is what
you need.

In your example, you have both an "or" (given) and an "and" (to be found),
so you use the "or" formula which relates them, and which does not require
the events to be independent or mutually exclusive:

   A = college graduate
   B = more than ten years experience

   P(A or B) = P(A) + P(B) - P(A and B)

         .75 = .56  + .53  -      x

Now solve for x.

You can't use the "and" formula to find P(A and B) because you don't know
that A and B are independent. It's also cagey thinking to notice that if
you did use it, you wouldn't be taking advantage of a piece of information
given in the problem: the "or" number. (In real life you can't use that
kind of savvy, but it helps when you're first learning.)

Also, there ARE two events here (two kinds of outcomes); that's not the
same as two EXPERIMENTS (such as picking two different people). It is true
that in simple problems, "and" tends to appear in compound events
involving two separate experiments (because that's the easiest way for two
events to be independent), whereas "or" often involves a single experiment
(e.g., picking one card that is either an ace or a spade). But you can
also have "and" problems involving independent events from only one
experiment (e.g., picking one card that is BOTH an ace AND a spade), and
"or" problems involving two experiments (e.g., picking two cards and
asking whether at least one is a spade -- though there's an easier way to
do this one than the "or" formula).

So you can't go by the number of selections, but only by whether the
events are independent, and whether you are looking for an "and" only, or
a relationship between "and" and "or."

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 08/21/2012 at 20:34:27
From: Millicent 
Subject: Thank you (Do mutually exclusive events always have to be an OR)

Thank you so much for you detailed explanation. It was extremely helpful
and now I clearly understand.



Date: 08/21/2012 at 20:39:41
From: Millicent 
Subject: Do mutually exclusive events always have to be an OR

Just a quick question that is sort of related to my earlier question: can
you still use P(A and B) = P(A) * P(B) for dependent events, like
P(picking 3 aces from a deck), or can you only use that for independent
events?



Date: 08/21/2012 at 22:31:00
From: Doctor Peterson
Subject: Re: Do mutually exclusive events always have to be an OR

Hi, Millicent.

I included in my list of rules one that you may not have seen, and which
is sometimes taught as if it were just P(A and B) = P(A) * P(B), but
applies even to independent events:

  If A and B are ANY events, then
  P(A and B) = P(A) * P(B | A)

Here P(B | A) is read as "the probability of B given A", and means that we
are finding the probability that B will happen, assuming that A has
happened. Sometimes this distinction is skipped over in introductory
classes, so they would write it as P(A and B) = P(A) * P(B).

In the example of picking two aces (to keep it as easy as possible to
write), you do this:

   P(two aces) = P(ace first and ace second)
               = P(ace first) * P(ace second | ace first)

Calculate "P(B)" not as it is initially, but as it is after picking the
first ace:

               = 4/52 * 3/51
               = 1/13 * 1/17 = 1/221

Again, the 3/51 isn't really the probability of picking an ace (the second
not being independent of the first), but is a conditional probability.

So with this caveat, you can use the product rule for dependent events.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
  


Date: 08/22/2012 at 03:46:50
From: Millicent 
Subject: Thank you (Do mutually exclusive events always have to be an OR)

Thank you for helping me again! It really helped, and I really appreciate
it.
Associated Topics:
High School Probability

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