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 questions involving the “at least” probability. 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 at least one vowel being picked?

- (A) 1/6

(B) 1/3

(C) 1/2

(D) 2/3

(E) 5/6

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

- (A) 5/8

(B) 13/16

(C) 15/16

(D) 31/32

(E) 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?

- (A) 1/2

(B) 9/16

(C) 11/16

(D) 13/16

(E) 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 “the probability 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) *Two dice are rolled. What is the probability of at least one of the dice rolling a 6? *

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) https://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.

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**

There are 14 women and 10 men on a committee. If three people are selected, what is the probability that:

a) They are all women?

b) There is at least one woman?

c) Only one man is chosen?

im studying for a test and absolutely bombing it I have so many questions god id drop my whole test review if I could im just not getting it no matter how much I put towards it if you could help me that would be amazing

Hi Daniel,

Sorry to hear you’re feeling so overwhelmed by the practice questions you’re doing. The good news is that I think I see the problem– the math question you’ve just shown me is

muchharder than any problem on the real GMAT would be. Actual GMAT Qunat problems deal with probability in a simpler, much more manageable way.This blog post shows more typical, GMAT-like probability questions involving “at least.” And this other GMAT probability questions article shows the typical structure and difficulty level of probability in general for GMAT Quant.

But, you may ask, how hard are the hardest GMAT Quant probability questions? We have a blog post for that too! See Mike’s set of advanced GMAT probability questions. That’s about as advanced as you need to get in order to do will with GMAT Quant probability.

Most importantly, you should always use the most realistic practice questions you can find, ones that resemble the real exam as much as possible. Studying for questions that are harder than the real exam may seem like a good way to bring your skills to the next level. But it’s actually harmful– you wind up studying math operations and question formats that are completely different from the ones on the actual test, leaving you unprepared on test day.

To find accurate, helpful GMAT practice problems, you can search this blog. You can also go for a Magoosh GMAT Premium subscription (or a free trial). And of course, you can get tons of official GMAT practice questions at the MBA.com store, including a wealth of free practice questions!

So make sure you’re studying realistically constructed GMAT practice questions. If you still feel overwhelmed by specific questions posted to this blog, you can always comment to ask for help. And if you do decide to subscribe to Magoosh GMAT, you’ll get email support for any Magoosh question and any official question. 🙂

There are 30 scratch-off tickets in one role and 6 of them are with monetary awards. If a lottery addict purchases two tickets, find the probability of:

a) both of them being lucky [3 marks]

b) at least one of them being lucky[4 marks]

c) exactly one of them being lucky [3 marks

how can we solve it?

Hi James!

If I’m understanding this correctly, there are 30 scratch-off tickets total, and 6 out of the 30 are “lucky”. Is that right? If not, let me know.

Assuming that the tickets do not get replaced, the probabilities will be as follows:

a) Both of them being lucky:6/30 * 5/29 = 30/870 = 3/87

b) At least one of them being luckyP(At least 1) = 1 – P(Neither are lucky)

P(Neither are lucky) = 24/30 * 23/29 = 92/145

1 – P = 53/145

c) Exactly one is luckyThere are two scenarios where you could get exactly one lucky ticket.

(Lucky)(Unlucky)

(Unlucky)(Lucky)

So, our probability is this:

2 * (6/30 * 24/29) = 48/145

Hope this helps! 😀

The table gives the score of a group of students in an English Language test.

Scores 2 3 4 5 6 7

No of Student 2 4 7 2 3 2

If a student is chosen at random from the group, what is the probability that he scored at least 6 marks

Please help

Hi Joshua,

This is actually a relatively easy one, because it only *looks* like the kind of “at least” question that Mike is discussing here. In this case, the use of “at least” is actually misleading. True “at least” questions involve multiple selections and multiple results: 2 selections of a letter, 6 flips of a coin, 3 draws of a card, etc…

However, the example problem you’ve given only involves one select/result. You merely have to choose one student from the overall student group.

So instead of the usual more complicated “at least” equation, all you need to do is recognize that if the student has a minimum score of 6, the student must come from one of two subgroups: either the student will be one of the 3 students who got a 6, or one of the 2 students who got a 7.

So there are 5 students in the group who got at least 6 marks. To find the possibility that one of these students will be selected from the whole group, simply divide 5 by the number of students in the whole group.

Whole group = 2+4+7+2+3+2 = 20

Subgroup that got 6 or higher on the test: 3+2 = 5

5/20 = 1/4

Possibility of selecting a student who scored 6 or higher is 1/4 aka 1 in 4, aka 25% chance. 🙂

Hi, in problem #3, why couldn’t we just calculate directly non-probabilty of winning in first two attempts? 3/4 +3/4*3/4

Hi Alex,

It’s great that you are thinking critically about this question! I’m not sure exactly what your thought process was in your setup here, but I can definitely explain an alternative method for getting the correct answer. The question is asking about the probability that we draw a heart on the third draw or later — it doesn’t matter when we draw a heart, just so long as it doesn’t happen on the first or second draw.

So instead of calculating the probability that we win in one or two draws and subtracting from one, we can DEFINITELY calculate the probability that you don’t draw a heart in the first two draws. Since the cards are replaced and there are 13 hearts in a 52 card deck, the probability of not drawing a heart will always be:

(52-13)/52 = 39/52 = 3/4

Thus, we can calculate the probability of not drawing a heart as:

3/4 * 3/4 = 9/16 – and that’s the answer! I hope that helps to clear up your doubt 🙂 It not, please provide me with some more information about why you set up the question in this way!

Thank you! That was exactly what I meant. In this problem the P(not A) = 1 – P(A) approach looks like overengineering…

“The question is asking about the probability that we draw a heart on the third draw or later — it doesn’t matter when we draw a heart, just so long as it doesn’t happen on the first or second draw” means what is the probability of drawing not-heart on first two attempts, which is 3/4×3/4=9/16.

Also, in your original explanation, why don’t you take into account the possibility of drawing hearts on first AND on the second attempt?

As I see it, it should be: P(win in two or fewer draws) = P(win in one draw OR win in two draws OR win in both, first and second draws). Or if we win on the first draw we don’t draw any more?

Thank you!

Hi again Alex,

I’m so glad this was helpful! I can understand how this might seem like “overengineering,” as you put it, but also consider how difficult it was to think through your method and set it up correctly. It seems like other students had similar problems in the comments, and it even took me a bit of head-scratching before I figured it out! At the end of the day, I’m not sure if this method is actually more efficient. The purpose of shortcuts like this is to provide you with clear paths forward–when you see “at least,” chances are that it will be more straightforward to use the complement rule. That being said, everyone ‘sees’ each question a bit differently, and you might find different path forwards that make more sense to you. So, I encourage you to continue to think critically about these questions and try to find multiple ways to solve them–this will allow you to find your own problem-solving style and build your number sense 🙂

For your second question, you answered it yourself 🙂 You use the OR probability because once you have drawn a heart, the game ends! You wouldn’t continue to draw after getting a heart, so you need to get a heart either in the first draw OR the second draw 😀

Hello,

Why is the probability of drawing a heart 1/4 instead of 1/13? I thought there were 13 cards in one suit, and so one suit of hearts is 13 total.

Thanks.

Hi Rosalyn,

Happy to help 🙂 The probability of drawing a heart from the deck is ¼ because there are 52 cards in the deck, and 13 of them are hearts. 13/52 = ¼.

You can also think of it this way: There are 4 suits in the deck, so the probability of drawing any one suit is always ¼.

There are 4 suits, 13 cards each, so 52 cards in a deck. So the probability of choosing one suit out of a deck is 13/52 (simplified as 1/4)! The fraction 1/13 doesn’t make sense because if you were asking about the probability of picking one heart out of 13 hearts, the probability would be 13/13.

Hello Mike,

in problem 3, can you explain why its “or” and not “and” for probability of not picking hearts the first and second time? so then it is picked the third time?

why isnt this right?

(1/4) * (1/4)= 1/16

1-(1/16) = 15/16 this is probability that at least 1 heart is not picked in the first two draws

1/4 * (15/16) = probability of first heart drawn on the 3 try

Hi Avni,

In this question, we are trying to find the probability of not drawing a heart until the third try. Put another way, this is 1-P(drawing a heard in the first two draws). We use an OR probability because it doesn’t matter in which try we draw the heart. It could be the first try OR the second try. It’s not an AND probability because we don’t need to draw two hearts–just one!

If we were to use (1/4)*(1/4), we are saying that a heart is picked in the first draw AND in the second draw. This is different than saying that we will win in two OR fewer draws–it is saying that we WILL win in BOTH draws, which is against the rules of the game!

G AND H ARE INDEPENDENT EVENTS. PROBABILITY THAT G WILL OCCUR IS r AND PROBABILITY THAT H WILL OCCUR IS s WHERE BOTH r AND s ARE GREATER THAN 0. THE PROBABILITY THAT EITHER G WILL OCCUR OR H WILL OCCUR BUT NOT BOTH IS ?? PLZ REPLY MR.MIKE

Hi Umer,

Happy to help 🙂 The probability that either G or H will occur but not both is

P(G or H) – P(G and H).P(G and H) is the product of P(G)*P(H).

And from the OR rule, we know

P(G or H) = P(G) + P(H) – P(G and H)

So,

P(G or H) – P(G and H) = P(G) + P(H) – P(G and H) – P(G and H)

= P(G) + P(H) – 2P(G and H)

= P(G) + P(H) – 2*P(G)*P(H)

And plugging in the values P(G) = r and P(H) = s, we have

P(G or H) – P(G and H) =

r + s – 2rsHope this helps 🙂

Hi Mike,

i have a question, actually 2 … could u explain as to how to do this? – i’m very weak in GMAT, would be of great help if you could let me know.

the probability of getting at least 3 heads when flipping a 2 sided coin 4 times?

the probability of getting 1 head when flipping a 2 headed coin 4 times?

Regards,

Joshua

the probability of getting at least 3 heads when flipping a 2 sided coin 4 times?

Ans:

P(x>3 head)= 1- p(x=0)+p(x=1)+p(x=2)

Solve this n u will get the answer. . Take 3 probibilities of x=0,1 n 2. Then subtract it from 1.

Mike, I can’t believe how comprehensive and articulate your posts are. This is high quality material and I can’t believe it’s free. Thank you for the effort you put into all the content – I’m especially grateful for the series of posts on probability.

Your posts are making a difference for gmat takers like me so keep up the great work… and keep crushing those fooseballs – I know I will! 🙂

Dave,

Thank you very much for the kind words, my friend. 🙂 I wish you the very best of luck in your GMAT preparations!

Mike 🙂

I know I am posting this in 2016, but I have to echo Dave’s sentiments. I have always struggled with combinatorics and probability – that is till I chanced upon your blog. This is indeed very high quality material – lucidly and comprehensively explained. A big thank you.

Example 3:

Not picking “Heart” one the first draw= 3/4

Not picking “Heart” one the second draw= 3/4

So, 1 -3/4*3/4 = 7/16

If Jessica rolls a die, what is the probability of getting at least a “3”?

i approached like this ,

not getting a 3=5/6

atleast=1-5\6

= 1/6

Hello Mike,

I still have problem how to incorporate the word “neither” i.e.

20% population catch flu

3% babies born (independently) have birth defect

consider two babies who

a) both have birth defects: = P(baby and baby) = P(baby) * P(baby) =03*0.03

b) neither baby catches flu ?????

When solving an “at least” problem, do you need to first ask yourself if the events are disjoint or not?

Melissa,

I’m happy to respond. 🙂 The “at least” question always arise in a scenario with repeated trials, in which the question is: how many times will such-and-such an event happen? Over these many trials, the question may ask, what is the probability that the event happens at least once? In any scenario of this sort, it seems to me that the various cases (

event doesn’t happen at all, event happens once, event happens twice,etc.) would always have to be disjoint. If they weren’t, there would have to be some kind of overlap between what it means for something to happen once and what it means to happen twice. I can’t think of any ordinary test-like scenario in which such an ambiguity or overlap would arise, so I would say these events are always disjoint.Does all this make sense? Do you have in mind some “at least” scenario in which it’s an issue whether something is disjoint?

Mike 🙂

No, I don’t have a scenario in mind. I was just trying to make myself a flow chart to help me decide how to approach all the different types of probability questions. Thank you for your answer!

Dear Melissa,

You are quite welcome. 🙂 I will say, go ahead and make a flowchart, but understand, a flowchart by itself is not necessarily going to lead to mastery of probability. Some areas of math, such as algebra, are very rule-based and methodically organizable; we can make clear step-by-step recipes for problem solving. Probability is very different: it’s all about pattern-recognition and learning how to frame scenarios correctly. It’s a much more right-brain area of math. For a discussion of related issues, see:

How to do GMAT Math Faster

I hope all this helps!

Mike 🙂

Nice mike!!!helped dis lot. ..I think it ll b more good if u add more problems with a problems at least 2…at least 3…

wow I like this page it help me to do my exam very well.

Dear Kori,

I’m glad you found this page helpful! Best of luck to you!

Mike 🙂

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.

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 #2P = (2/5)*(5/6) = 2/6 = 1/3

Case two:pick a vowel from Set #2, and not from Set #1P = (3/5)*(1/6) = 1/10

Case three:pick a vowel from bothP = (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

NOTtrue 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 🙂

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 ?

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 whatis 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 🙂

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.

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

. 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 theselectionnextroll.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 🙂

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

Dear Pawan,

Yes, that’s a very efficient way to calculate the answer. Best of luck to you, my friend.

Mike 🙂

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..

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 seePawanwrote: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

the opposite case of “at least three draws to win” — in fact, it’s another way of sayingnotexactly 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

Pawansaid. In fact, it is not correct at all, and the explanation above shows the only correct solution.Does all this make sense?

Mike 🙂

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?

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 🙂

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

Dear Andrew,

You are quite welcome. Best of luck to you!

Mike 🙂

what if there is “at least 2 “, ” exactly”, “atmost 2 out of 5” in Questin

Dear Fida,

This blog article:

https://magoosh.com/gmat/2014/gmat-advanced-probability-problems/

might answer some of your questions.

Mike 🙂

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

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 🙂

Hi Mike,

I too thought the same way. But I added an extra 1/4 (the probability of drawing the heart) at the end:

Probability of winning in the 3rd draw = (3/4)*(3/4)*(1/4) =9/64

Why am I wrong?

Hi Vishal,

The problem with your calculation is that it is “the probability of winning in EXACTLY the 3rd draw.” However, we were asked, “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?”

This means that the question is looking for the probability that the heart is drawn on exactly on the third draw, exactly on the fourth, exactly on the fifth, or so on. We are not only looking at the third draw.

Oh Okay. I realize it now.

Thank you very much. All these blogs are wonderful. Clear and explained in detail 🙂