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GMAT Math: Probability Rules

First, some practice questions.  The scenario below is relevant to questions #1-#3.

There are two sets of letters, and you are going to pick exactly one letter from each set.

Set #1 = {A, B, C, D, E}

Set #2 = {K, L, M, N, O, P}

1) What is the probability of picking a C and an M?

  1. 1/30
  2. 1/15
  3. 1/6
  4. 1/5
  5. 1/3

2) What is the probability of picking a C or an M?

  1. 1/30
  2. 1/15
  3. 1/6
  4. 1/5
  5. 1/3

3) What is the probability of picking two vowels?

  1. 1/30
  2. 1/15
  3. 1/6
  4. 1/5
  5. 1/3

_______________________________________________________________________________

4) 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 and is female?

  1. 1/20
  2. 1/10
  3. 1/5
  4. 2/5
  5. 3/4

5) 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?

  1. 1/20
  2. 1/10
  3. 1/5
  4. 2/5
  5. 3/4

 

The simplistic probability rules

Here is the absolute bare minimum you need to know for probability calculations on the GMAT:

“AND” means MULTIPLY

“OR” means ADD

Is this the whole story? Well, not exactly.  But if you can’t remember or don’t understand anything else about probability, at least know these two bare-bones rules, because just this will put ahead of so many people.  Just this is enough to solve the problems #1 and #3, although these alone could lead to problems on the others.  Before we qualify these simplistic rules, we need to discuss two distinctions.

 

Disjoint

Two events are disjoint if they are mutually exclusive.  In other words, two events are disjoint if the probability of their simultaneous occurrence is zero, that is, it is absolutely impossible to have them both happen at the same time.   For example, different faces of a single die are disjoint: under ordinary circumstances, if you roll one die once, you can’t simultaneously get, say, both a 3 and a 5.   Those two are disjoint.  Suppose we are picking random people and classifying them by their current age.  In this process, being in the category “teenager” and being in the category “senior citizen” are disjoint: there is no one we could pick who is simultaneously in both categories.   Most categories involving human beings are too messy for the distinction “disjoint” to apply.

If events A and B are disjoint, then we can use the simplified OR rule:

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

That’s the case in which the simplified rule, OR means ADD, works perfectly.  If events A and B are not disjoint, then we have to use the generalized OR rule:

P(A or B) = P(A) + P(B) – P(A and B)

The reason for that final term: we need to subtract the overlap.  The events in the region “A and B” are included in region A and also in region B, so if we add those two regions, the overlap gets counted twice.  We need to subtract it, so it is only counted once like everything else.

 

Independent

Two events are independent if whether one happens has absolutely no influence on whether the other happens.  In other words, knowing about the outcome of one event gives absolutely no information about how the other event will turn out.   For example, if I roll two ordinary dice, the outcome of each die is independent of the other die.  If I tell you I rolled two dice, and the first die was a 4, then knowing that give you no clue about what the number on the other die might be.   On any given day, what the weather is in the San Francisco Bay Area and how the Dow Jones Industrial Average performs are independent: knowing one gives us absolutely no information about how the other turned out.

We have to be careful.  If I shuffle a deck of cards, draw one, replace it, re-shuffle, and draw another, then the two cards are independent.  BUT, if I shuffle the deck, draw one card, and then without replacement draw a second card, then they are not independent.  For example, if the first card is the 7 of Hearts, then it is less likely that the second card would be either a 7 or a Heart, because there are fewer of those options among the remaining 51 cards.

Also, notice that there are many human situations which would be independent in a perfect just ideal world, but regrettably are not independent in a real world full of inequities.   In a perfect world, gender and corporate promotion would be independent, but in practice, they are not.  In a perfect world, race and criminal conviction would be independent, but in practice, they are not.

If events A and B are independent, then we can use the simplified AND rule:

P(A and B) = P(A)*P(B)

That’s the case in which the simplified rule, AND means MULTIPLY, works perfectly.  If events A and B are not independent, then things get complicated.  Technically, the “generalized AND rule” formula would involve a concept known as “conditional probability“, which would lead into realms of probability theory that are  tested less frequently on the GMAT.  See that other blog that discusses conditional probability if you want to understand this advanced topic in more detail.

 

Practice

Having read this post, take another look at those practice questions, and see if you understand them better, before simply reading the explanations below.   Be patient with yourself as you work through probability: it takes time to internalize these distinctions, such as “disjoint” and “independent”.  There will be more information in the next post in this sequence.

 

Practice problem solutions

1) Whatever we pick from the first set is independent with whatever we pick from the second set, so we can use the simplified AND rule.

P(first pick = C) = 1/5

P(second pick = M) = 1/6

P(C and M) = P(C)*P(M) = (1/5)*(1/6) = 1/30

Answer = A

2) Picking an M is not disjoint with picking a C — they both could happen on the same round of the game.   We have to use the generalized OR rule for this:

P(C or M) = P(C) + P(M) – P(C and M)

Fortunately, we know the first two, and we calculated the value of the third term already in #1.

P(C or M) = P(C) + P(M) – P(C and M)

Answer = E

3) On the first pick, two of the five letters are vowels — A & E — so the probability of picking a vowel on the first pick is 2/5.   On the second pick, only one letter out of the six is a vowel — O — so the probability of picking a vowel on the second pick is 1/6.  The two picks are independent: what one selects from one set has absolutely no bearing on what one picks from the other set.   Therefore, we can use the generalized AND rule.

P(two vowels) = P(vowel on first pick)*P(vowel on second pick) =(2/5)*(1/6) = 2/30 = 1/15

Answer = B

4) Here we have an AND question, and the parameters — gender and advanced degree — are not independent.  If I tell you the gender of a certain employee, then that give me information about how likely it is that this employee has an advanced degree.   One parameter gives information about the other, which means they are not independent.  Therefore, we cannot use the simplified AND rule.  Fortunately, it is relatively easy here to calculate everything directly.

There are 100 female employees, and we know 40% of them have advanced degrees, so there are 40 employees who both are female and have an advanced degree.  That’s the number of employees in the AND region.  Well, there are 400 employees altogether.  Of these 400 total employees, the probability of picking someone in this AND region is

P = 40/400 = 1/10

Answer = B

5)  In this corporation, there are 400 total employees.  There are 100 women.  Of the 300 men, 20% have advanced degrees —-10% of 300 must be 30, so 20% of 300 must be 60.  Add the women and the men with advanced degrees: 100 + 60 = 160.   This is the OR region, full set of individuals that satisfy the condition “has an advanced degree or is female.”  Of the 400 employees, what’s the probability of picking one of the 160 in this particular group?

P = 160/400 = 16/40 = 4/10 = 2/5

Answer = D

Try out some more GMAT probability problems here.

About the Author

Mike McGarry is a Content Developer for Magoosh with over 20 years of teaching experience and a BS in Physics and an MA in Religion, both from Harvard. He enjoys hitting foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Follow him on Google+!

20 Responses to GMAT Math: Probability Rules

  1. RD April 7, 2014 at 7:50 am #

    Mike in Question 4,

    should it not be just 40/100

    Total no of female students is 100 and the number with advanced degrees is 40.

    If not are the 2 questions different?

    1.What is the probability that a student is female an has an advanced degree

    2. What is the probability that one of the students with advanced degree is a female?

    Thanks

    • Mike
      Mike April 7, 2014 at 12:14 pm #

      Dear RD,
      Question #4 asks: “If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree and is female?” You are correct that the 40 females with advanced degrees are the “desired” result, the numerator of the probability fraction, but the choice is not made from all female employees (100) — instead, it is made from all 400 employees, so that’s the proper denominator. 40/400 = 1/10. In probability questions, one always has to read carefully, to make sure that one both choosing the right group and choosing from the right group.
      Probability questions have to be phrased extremely carefully. Your second question is an intriguing question — of all the people (male & females) with advanced degrees, how many are females. That’s a question that wasn’t asked here. 60 males and 40 females have advanced degrees, so of the 100 employees with advanced degrees, 40$ or 4/10 are females.
      Your first question is poorly worded. It seems to mean the same as question #4 here, but that wording would need to be cleaned up to be a GMAT worthy question. It’s not an easy thing to write a perfectly clear and unambiguous GMAT probability question.
      Mike :-)

  2. Dude March 12, 2014 at 12:25 pm #

    P (a) + P (b) – P (a And b)

    1/4 + 1/4 – 1/10 = 2/5

    Why is the answer not written as it was just taught above?

    • Mike
      Mike March 12, 2014 at 1:37 pm #

      Dude,
      As is often the case with probability, there is more than one way to think about and solve a problem. Once we calculate P(A) and P(B), the formula you suggest becomes another valid solution. It’s good to see more than one way to solve a problem.
      Mike :-)

  3. Heather January 15, 2014 at 9:40 pm #

    Hi Mike,

    I’m confused with question #2. Why are the two sets not mutually exclusive, thus making the probability of C OR M (11/30). Why do we need to subtract the P (C and M)?

    Thanks!
    Heather

    • Mike
      Mike January 16, 2014 at 10:21 am #

      Heather,
      That’s a fantastic question. In the process of picking two letters, we wind up with a set of two, a pair. In that pair, it’s possible for C and M to happen at the same time — the result of “get C as a member of the pair” is not mutually exclusive with the result “get M as a member of the pair.” The probability question in #2 is about the pair that results, so we need to consider the pair as a whole, not the individual selections in isolation.
      Does this make sense?
      Mike :-)

  4. Jen January 4, 2014 at 12:01 pm #

    Hi,

    I’m confused about the Independent section.
    In the previous section, you said “If events A and B are not disjoint, then we have to use the generalized OR rule: P(A or B) = P(A) + P(B) – P(A and B)”

    And then you said “If events A and B are not disjoint, then things get complicated.”
    So does “complicated” mean using the generalized OR rule?

    Also, probability, permutation, combination and counting really confuse me. Do you have any advice in which order of your blogs I should read?

  5. Mihaela October 1, 2013 at 1:52 am #

    Dear Mike,
    In the problem # 5, in the explanation section, there are 100 employees with an advanced degree, isn’t it? Therefore, the solution still remains 2/5?

    • Mike
      Mike October 1, 2013 at 10:02 am #

      Dear Mihaela,
      The question asks: “What is the probability this employee has an advanced degree OR is female?” If the question simply asked “What is the probability this employee has an advanced degree?”, then that would be the 100 folks with advanced degrees over 400 total, p = 0.25. BUT, we have an OR question. The number of people with advanced degrees, 100, is tricky because it includes both men & woman, and if we also include all 100 women, then those woman with advanced degrees are in the overlap region and get counted twice. When we count them only once, we get men with advanced degrees (60) + ALL women (100) = 160 — 160/400 = 16/40 = 2/5.
      Does all this make sense?
      Mike :-)

      • Mihaela October 1, 2013 at 10:36 am #

        Thank you for the reply!! Yes, this makes sense! I was not paying enough attention to your explanations! I

        M.

        • Mike
          Mike October 1, 2013 at 10:50 am #

          Mihaela,
          You are more than welcome. Best of luck to you!
          Mike :-)

  6. LC September 25, 2013 at 7:29 am #

    #5…I was confused with the solution as well….but my mistake was adding the overlap of women with adv degrees into my solution.

    I must tell myself…DON’T DOUBLE DIP…as in….the solution for #5 is correct because we want = ALL WOMEN + ALL ADV DEGREES – DOUBLE DIPPERS…..equivalent to ALL WOMEN + (ALL ADV DEGREE – WOMEN w/ ADV DEGREE) > ALL WOMEN + MEN w/ ADV DEGREE.

    Thanks for the challenging questions. This has been helpful!

    • Mike
      Mike September 25, 2013 at 10:25 am #

      Dear LC,
      That’s a good way to say it — double-dipping is problematic in this context, as it is in many others!! :-) I’m glad you found this helpful. Best of luck to you!
      Mike :-)

  7. Karan April 11, 2013 at 10:18 pm #

    In question number 5. “Advanced degree or Female”. In this case these two events are mutually exclusive? Because one can be someone who is an advanced degree holder as well as a female.

    • Mike
      Mike April 12, 2013 at 11:00 am #

      Karan,
      That question explicitly says: “40% of the females have advanced degrees” — if there’s even one person who’s in both categories, female and advanced degree, that means they are NOT mutually exclusive. —– As a general rule, “mutually exclusive” applies to things like dice or playing cards, not to humans. Humans & categories of any sort are always messy and complicated, not pure & simple like dice.
      Mike :-)

      • KC June 8, 2013 at 7:48 am #

        Hi,

        I am also confused with the problem 5. Why are we not subtracting the case of both “advanced degree” and “female”? It seems like the probability of 160/400 includes females who have advanced degree, which we do not want. I thought we only want females who do not have advanced degree or males who have advanced degree.

        Can you explain it to me what are some differences of problem 2 and 5?

        Thank you!

        KC

        • Mike
          Mike June 10, 2013 at 10:09 am #

          KC,
          In logic, there’s a distinction between “inclusive OR”, which means that “A or B” includes A by itself, B by itself, and A & B together, and “exclusive OR”, which includes A by itself and B by itself but not A & B together. In GMAT math problems, whenever the word “or” appears, you must assume that it is an “inclusive OR”. If what they mean is an “exclusive or”, they would have to specify that explicitly.
          There are a number of important differences between #2 & #5, the most significant of which is — in #2, there are two separate choices, a choice from Set #1 and a choice from Set #2. In #5, we are just choosing one person with one of two qualities. Very different.
          Mike :-)


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