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Probability Problem-

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by baalok88 » Thu Nov 03, 2016 9:35 am
In a certain corporation, there are 300 male employees and 100 female employees. It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree or is female?

(A) 1/20
(B) 1/10
(C) 1/5
(D) 2/5
(E) 3/4

----My approach---
Males with Adv.Degree= 60
Females with Adv. Degree= 40
Probability of Advance degree (100/400) Or Females (100/400) = 100/400+ 100/400= 1/2

Can anybody explain me what did I do wrong that the answer choice does not have 1/2 as an option?
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by DavidG@VeritasPrep » Thu Nov 03, 2016 10:23 am
baalok88 wrote:In a certain corporation, there are 300 male employees and 100 female employees. It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree or is female?

(A) 1/20
(B) 1/10
(C) 1/5
(D) 2/5
(E) 3/4

----My approach---
Males with Adv.Degree= 60
Females with Adv. Degree= 40
Probability of Advance degree (100/400) Or Females (100/400) = 100/400+ 100/400= 1/2

Can anybody explain me what did I do wrong that the answer choice does not have 1/2 as an option?
The problem is that "females with advanced degrees" are included in both the 'female" group and the "Advanced Degree" group. So when we add those two groups together, we're double-counting the overlap. (Think of the overlap in a Venn diagram.) So there are 100 females, 100 people total with advanced degrees, and 40 females with advanced degrees.

To make sure we don't double-count the females with advanced degrees, we'll subtract the overlap: 100 + 100 - 40 = 160.

160/400 = 2/5

P(female or advanced degree) = 160/400 = 16/40 = 2/5. Answer is D
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by DavidG@VeritasPrep » Thu Nov 03, 2016 10:26 am
Alternatively, you can think of your female OR advanced degree population as divided into three groups.

1) Females with an adv degree
2) Females without an adv degree
3) Males with an adv degree

1) You correctly found this is 40
2) If there are 100 females, and 40 have the degree, then 100-40 = 60 females don't have it
3) You correctly found this is 60

Sum them up: 40 + 60 + 60 = 160 are female OR have an advanced degree. 160/400 = 2/5, or D
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by fiza gupta » Fri Nov 04, 2016 9:49 am
Total male employees: 300
20% have advanced degrees : 300*.2 = 60
Total female employees: 100
40% have advanced degrees : 100*.4 = 40
total students having advanced degrees : 100

females + advanced degrees = 100 + 100 - 40(double count of females having advanced degrees)
= 160

= 160/400 = 2/5

SO D
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by Scott@TargetTestPrep » Fri Nov 04, 2016 3:02 pm
baalok88 wrote:In a certain corporation, there are 300 male employees and 100 female employees. It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree or is female?

(A) 1/20
(B) 1/10
(C) 1/5
(D) 2/5
(E) 3/4
We are given that there are 400 total employees - 100 female and 300 male. We are also given that 20% of the male employees have advanced degree and that 40% of the females have advanced degrees.

Thus,

0.2 x 300 = 60 males with advanced degrees

And

0.4 x 100 = 40 females with advanced degrees

We see that 100 total employees had an advanced degree.

Since there are females with advanced degrees we have to subtract out any overlap when determining the probability of selecting a female or an employee with an advanced degree.

P(female or advanced degree) = P(female) + P(advanced degree) - P(both)

P(female or advanced degree) = 100/400 + 100/400 - 40/400

P(female or advanced degree) = 160/400 = 16/40 = 2/5

Answer: D

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by AndyMichael89 » Sat Nov 05, 2016 3:38 am
Hi everyone,

I understood the way how to solve this one but I am just not sure about the general rule that needs to be applied.
In one book I found the following formula: P(A or B)= P(A) + P(B) - P(A and B) [General Addition Rule]

However, according to that one it would be: 1/4 + 1/4 - (1/4 x 1/4) which would lead to a wrong result. Could you please help me out?

Greetings!
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by [email protected] » Sat Nov 05, 2016 9:24 am
Hi AndyMichael89,

To start, I'm not sure where you're getting your numbers (I assume that you're just using 1/4s as a basic example). The formula that you're citing defines the probability of "A OR B but NOT both." In this question, we're actually looking for the probability of "A or B or BOTH."

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by Matt@VeritasPrep » Fri Nov 11, 2016 3:41 pm
AndyMichael89 wrote: In one book I found the following formula: P(A or B)= P(A) + P(B) - P(A and B) [General Addition Rule]
Also worth noting that this is easiest to use in the form given if A and B are independent. That will typically be true on the GMAT, but not always.
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by Matt@VeritasPrep » Fri Nov 11, 2016 3:42 pm
AndyMichael89 wrote: In one book I found the following formula: P(A or B)= P(A) + P(B) - P(A and B) [General Addition Rule]
And, of course, I should mention the mathematical definition of "or", which drives students crazy. "A or B" in everyday life means "exactly one of A or B", but in math and logic, "A or B" means "at least one of A and B"! This is a crucial difference, but one that's easy to overlook, forget, or never have learned.
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by AndyMichael89 » Sun Nov 13, 2016 4:36 am
Dear Rich,

thank you for your explanation so far.

So it all comes down to the "exclusive or" - "inclusive or" thing, right?

For instance, if we have 100 school kids and 20 have blond hair (A) and 40 are female (B) there could be two scenarios, right?:
In the first one we would use the formular P(A or B) = P(A) + P(B) - P(both)
In the second one we would use the formular P(A or B) = P(A) + P(B) - P(A and B)

Could you please explain when I would use which formula based on my example regarding the wording of the question?

Sorry, if I misunderstood this topic - I cannot quite get my head around this ...
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by Matt@VeritasPrep » Fri Nov 25, 2016 3:56 pm
AndyMichael89 wrote: So it all comes down to the "exclusive or" - "inclusive or" thing, right?
But in math and logic there is no "exclusive or", at least not in those terms. If you see "or", it is ALWAYS INCLUSIVE. To state the exclusive, you'd need something like "exactly one of x and y".
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