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GMAT Math: the Probability “At Least” Question

In the first post in this series, I spoke about the AND rule and the OR rule in probability.  Now, we will focus on probability question involve the words “at least.”  First, some practice of this genre.

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

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

1) There are these two sets of letters, and you are going to pick exactly one letter from each set.  What is the probability of picking at least one vowel?

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

2) Suppose you flip a fair coin six times.  What is the probability that, in six flips, you get at least one head?

  1. 5/8
  2. 13/16
  3. 15/16
  4. 31/32
  5. 63/64

3) In a certain game, you pick a card from a standard deck of 52 cards.  If the card is a heart, you win.  If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again.  The person keeps repeating that process until he picks a heart, and the point is to measure: how many draws did it take before the person picked a heart and won?   What is the probability that one will have at least two “heartless” draws on the first two draws, not picking the first heart until at least the third draw?

  1. 1/2
  2. 9/16
  3. 11/16
  4. 13/16
  5. 15/16

 

The complement rule

There is a very simple and very important rule relating P(A) and P(not A), linking the probability of any event happening with the probability of that same event not happening.   For any well-defined event, it’s 100% true that either the event happens or it doesn’t happen. The GMAT will not ask you probability question about bizarre events in which, for example, you can’t tell whether or not the event happened, or complex events which could, in some sense, both happen and not happen.  For any event A in a probability question on the GMAT, the two scenarios “A happens” and “A doesn’t happen” exhaust the possibilities that could take place.  With certainty, we can say: one of those two will occur.  In other words

P(A OR not A) = 1

Having a probability of 1 means guaranteed certainty.  Obviously, for a variety of deep logical reasons, the events “A” and “not A” are disjoint and have no overlap.  The OR rule, discussed in the last post, implies:

P(A) + P(not A) = 1

Subtract the first term to isolate P(not A).

P(not A) = 1 – P(A)

That is known in probability as the complement rule, because the probabilistic region in which an event doesn’t occur complements the region in which it does occur.   This is a crucial idea in general, for all GMAT probability questions, and one that will be very important in solving “at least” questions in particular.

 

The complement of “at least” statements

Suppose event A is a statement involving words “at least” — how would we state what constituted “not A”?  In other words, how do we negate an “at least” statement?  Let’s be concrete.  Suppose there is some event that involves just two outcomes: success and failure.  The event could be, for example, making a basketball free throw, or flipping a coin and getting heads.  Now, suppose we have a “contest” involving ten of these events in a row, and we are counting the number of successes in these ten trials.   Let A be the event defined as: A = “there are at least 4 successes in these ten trials.”  What outcomes would constitute “not A”?  Well, let’s think about it.  In ten trials, one could get zero successes, exactly one success, exactly two successes, all the way up to ten successes.  There are eleven possible outcomes, the numbers from 0 – 10, for the number of successes one could get in 10 trials.  Consider the following diagram of the number of possible successes in ten trials.

The purple numbers are the members of A, the members of “at least 4 successes” in ten trials.  Therefore, the green numbers are the complement space, the region of “not A.”  In words, how would we describe the conditions that land you in the green region?  We would say: “not A” = “three or fewer success” in ten trials.   The negation, the opposite, of “at least four” is “three or fewer.”

Abstracting from this, the negation or opposite of “at least n” is the condition “(n – 1) or fewer.”  One particularly interesting case of this is n = 1: the negation or opposite of “at least one” is “none.”   That last statement is a hugely important idea, arguably the key to solving most of the “at least” questions you will see on the GMAT.

 

Solving an “at least” question

The big idea for any “at least” question on the GMAT is: it is always easier to figure out the complement probability.  For example, in the above scenario of ten trials of some sort, calculating “at least 4″ directly would involve seven different calculations (for the cases from 4 to 10), whereas the calculation of “three or fewer” would involve only four separate calculations (for the cases from 0 to 3).  In the extreme — and extremely common —- case of “at least one”, the direct approach would involve a calculation for almost case, but the complement calculation simply involves calculating the probability for the “none” case, and then subtracting from one.

P(not A) = 1 – P(A)

P(at least one success) = 1 – P(no successes)

This is one of the most powerful time-saving shortcuts on the entire GMAT.

 

An example calculation

Consider the following simple question.

4) A two dice are rolled.  What is the probability of rolling a 6 on at least one of them? 

It turns out, calculating that directly would involve a relatively long calculation — the probability of exactly one 6, on either die, and the rare probability of both coming up 6’s.  That calculation easily could take several minutes.

Instead, we will use the shortcut defined above:

P(not A) = 1 – P(A)

P(at least one 6) = 1 – P(no 6’s)

What’s the probability of both dice coming up no 6’s?  Well, first, let’s consider one die. The probability of rolling a 6 is 1/6, so the probability of rolling something other than 6 (“not 6″) is 5/6.

P(two rolls, no 6’s) = P(“not 6″ on dice #1 AND “not 6″ on dice #2)

As we found in the previous post, the word AND means multiply.  (Clearly, the outcome of each die is independent of the other).   Thus:

P(two rolls, no 6’s) =(5/6)*(5/6) = 25/36

P(at least one 6) = 1 – P(no 6’s) = 1 – 25/36 = 11/36

What could have been a long calculation becomes remarkably straightforward by means of this shortcut.   This can be an enormous time-saver on the GMAT!

 

Practice

Having read this post, you may want to take another shot at the three practice questions above before reading the solutions below.  Also, here’s a free question, with video explanation, on this same topic:

5) http://gmat.magoosh.com/questions/839

The next article in this series will explore probability questions that involve counting techniques.

Practice problem explanations

1) P(at least one vowel) = 1 – P(no vowels)

The probability of picking no vowel from the first set is 3/5.  The probability of picking no vowel from the second set is 5/6.  In order to get no vowels at all, we need no vowels from the first set AND no vowels from the second set.  According to the AND rule, we multiply those probabilities.

P(no vowels) = (3/5)*(5/6) = 1/2

P(at least one vowel) = 1 – P(no vowels) = 1 – 1/2 = 1/2

Answer = C

2)  P(at least one H) = 1 – P(no H’s)

In one flip, P(“not H”) = P(T) = 1/2.   We would need to have this happen six times — that is to say, six independent events joined by AND, which means they are multiplied together.


Answer = E

3) A full deck of 52 cards contains 13 cards from each of the four suits.  The probability of drawing a heart from a full deck is 1/4.  Therefore, the probability of “not heart” is 3/4.

P(at least three draws to win) = 1 – P(win in two or fewer draws)

Furthermore,

P(win in two or fewer draws) = P(win in one draw OR win in two draws)

= P(win in one draw) + P(win in two draws)

Winning in one draw means: I select one card from a full deck, and it turns out to be a heart.  Above, we already said: the probability of this is 1/4.

P(win in one draw) = 1/4

Winning in two draws means: my first draw is “not heart”, P = 3/4, AND the second draw is a heart, P = 1/4.  Because we replace and re-shuffle, the draws are independent, so the AND means multiply.

P(win in two draws) =(3/4)*(1/4) = 3/16

P(win in two or fewer draws) =P(win in one draw) + P(win in two draws)

= 1/4 + 3/16 = 7/16

P(at least three draws to win) = 1 – P(win in two or fewer draws)

= 1 – 7/16 = 9/16

Answer = B

 

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+!

18 Responses to GMAT Math: the Probability “At Least” Question

  1. BBQ December 11, 2014 at 7:56 am #

    Hi Mike,
    could you please provide the “long” solution for the first question so I could understand the problem more deeply? Now I am not exactly sure why that kind of solution solution ((2/5)*(1/6)=1/15) doesn’t work. I tried to solve the problem using another (longer) approach but not sure where to start.

    • Mike
      Mike December 11, 2014 at 1:47 pm #

      Dear BBQ,
      My friend, often we do not answer questions that require us to explain a lot beyond what we already explained in the blog, and I am not sure that seeing the other cases will help you, but maybe this will. Because it’s quick, I will list these out:
      Case one: pick a vowel from Set #1, and not from Set #2
      P = (2/5)*(5/6) = 2/6 = 1/3
      Case two: pick a vowel from Set #2, and not from Set #1
      P = (3/5)*(1/6) = 1/10
      Case three:pick a vowel from both
      P = (2/5)*(1/6) = 1/15
      The outcome of “at least one vowel” includes all three of these cases, we have to add those probabilities.
      1/3 + 1/10+ 1/15 = 10/30 + 3/30 + 2/30 = 15/30 = 1/2
      That’s a much longer calculation that the complement rule calculation I demonstrated in the solution.
      My friend, I believe your mistake was to assume that the complement of “neither” was “both”. “Both” and “neither” are NOT true opposites, not true complements, because together, they don’t include all possibilities. The two cases of “yes” from one set and “no” from the other, cases one & two here, have to be included — both of these cases must be part of the complement of either “both” or “neither.”
      Similarly, people naively assume that the “all” and “none” are opposites, and they are not. The opposite of the false statement “All human beings a baseball fans” is the true statement “Some human beings are not baseball fans.” The opposite of the false statement “No human beings are baseball fans” is the true statement “Some human beings are baseball fan.” Both the “all” and the “none” statements are false, so they can’t be opposites of each other.
      Does all this make sense?
      Mike :-)

  2. Vijay August 18, 2014 at 7:49 pm #

    Hi Mike,

    If I try solving the first problem using Ven Diagrams, Set1 {A, B, C, D, E} and Set 2 {K, L, M, N, O, P} have no intersections, so can we not say they are mutually exclusive ?

    • Mike
      Mike August 18, 2014 at 10:41 pm #

      Dear Vijay,
      I’m happy to respond. :-) My friend, this is part of what is SO TRICKY about probability. You have to look at everything the right way and frame it the right way. Yes, the two sets do not overlap, but that’s completely irrelevant, because whether something comes from one set or the other is not a matter of chance. By design, we have decided at the outset that we definitely will pick one letter from Set #1 and one from Set #2, so that’s fixed: that’s not where the probability is. Probability simply does not concern things that are fixed in advance. The probability concerns the letters chosen. If I pick, say, an A from the first set, is that mutually exclusive with any letter I could pick from the second set? That’s the question you need to ask.
      You have to think very carefully about exactly what is being chosen. Elements of the problem that are fixed by design are not where the probability is happening — probability very specifically concerns those elements that can vary each time.
      Does all this make sense?
      Mike :-)

      • ogoma December 3, 2014 at 10:36 am #

        Hi Mike,

        If I pick, say, an A from the first set, is that mutually exclusive with any letter I could pick from the second set? My answer is Yes. I don’t understand why your answer is otherwise. Please clarify.

        Thanks.

        • Mike
          Mike December 3, 2014 at 11:39 am #

          Dear Ogoma,
          I think some confusion is arising about the exact meaning of “mutually exclusive.” We are NOT talking about sets and which sets overlap. Clearly, the first set of letters contains none of the letters in the second set, and vice versa, so we could say that, as sets, these two sets are “mutually exclusive,” but this is NOT what is meant by the term in Probability — in fact, it’s almost the opposite of what is meant in Probability!
          In Probability, our concern is selection. If I pick one option, and picking this prevents me from picking another option, then the two options are mutually exclusive. When I roll a single die once, getting an odd number and getting an even number are two mutually exclusive events, because no number I get on the single roll could possibly be both at the same time — getting an even number, by definition, completely excludes the possibility that the same number is also odd. By contrast, getting an even number and getting a prime number are not mutually exclusive, because 2 is both at the same time. Similarly, getting an odd number on one roll of a die is not at all mutually exclusive with getting an even number on the next roll.
          In this scenario, “mutually exclusive” is all about what letter we have picked and what we are thereby able to pick. If I select A on my first choice, does the very act of choosing that letter automatically prevent me from choosing any of the members of the second set? No! Once I pick A on my first choice, I could freely pick any member at all from the second set on my second choice. Nothing is excluded in the act of selection. Therefore, in terms of Probability, A is not mutually exclusive with any element of the second set.
          Does all this make sense?
          Mike :-)

  3. pawan August 9, 2014 at 1:31 am #

    HI Mike

    for question 3, Can I directly write this as

    P(at least three draws to win) = 1 – P(not win in 2 draws)

    1-(3/4 * 3/4) = 1-9/16 = 7/16

    • Mike
      Mike August 9, 2014 at 3:26 pm #

      Dear Pawan,
      Yes, that’s a very efficient way to calculate the answer. Best of luck to you, my friend.
      Mike :-)

      • Julia December 11, 2014 at 7:32 am #

        I also thought so. But we don’t have the answer 7/16 among the answer choices and according to Mike’s explanations the answer is 9/16. Confusing..

        • Mike
          Mike December 11, 2014 at 1:34 pm #

          Julia,
          I’m happy to respond! First of all, very good eye! Yes, I think I glanced at Pawan‘s work too quickly. Yes, this is possible to approach with an “at least” solution: that’s what I show in the solutions above. Now that I look more carefully, I see Pawan wrote:
          P(at least three draws to win) = 1 – P(not win in 2 draws)
          Technically, that is nonsense, because “not winning in 2 draw” is not the opposite case of “at least three draws to win” — in fact, it’s another way of saying exactly the same thing. Subtracting the opposite is not about saying the exactly same thing in a different way. The true opposite, the true complement, of “at least three draws to win” is “winning in one of first two draws.” Those are two outcomes that truly do not overlap or include each other in any way. That’s what it means to be a complement in Probability.
          So, I am sorry that I made the mistake of initially approving what Pawan said. In fact, it is not correct at all, and the explanation above shows the only correct solution.
          Does all this make sense?
          Mike :-)

  4. Andrew May 13, 2014 at 3:58 pm #

    For question 3, am I the only one that interpreted “What is the probability that one will have at least three draws before one picks a heart?” as “What is the probability that one will have at least three draws before drawing a heart”? What I’m getting at is that if drawing is the same as picking, which it seems to be, then the probability “at least” region and complement region ought to be draws 4 through 10 and 1 through 3, respectively. Right?

    • Mike
      Mike May 14, 2014 at 1:57 pm #

      Andrew,
      I’m happy to respond. :-)
      You are perfectly correct: the wording there left some ambiguity, and could have been read in more than one way. Sometimes it hard to write something precisely, with one interpretation in mind, and automatically be able to see the way that someone else might interpret it differently. I just rewrote that sentence, making it considerably more wordy, but I think it’s clear now. Does it look clear to you?
      Mike :-)

      • Andrew May 14, 2014 at 9:40 pm #

        Yes it does, thanks for clarifying that for me. Your blog is great and has helped me tremendously.

  5. vittovangind April 27, 2014 at 7:52 am #

    For exercise #3 above, can’t we simply reformulate as “what is the probability of not drawing a heart in the first 2 rows:

    P(no heart in draw #1) and P(no heart in draw #2) = 3/4 * 3/4 = 9/16

    • Mike
      Mike April 28, 2014 at 9:56 am #

      Dear Vittovangind,
      Yes, that is another perfectly valid way of thinking about it. Here that works. It’s always good to see more than one solution.
      Mike :-)


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