If you’re like most students, you probably struggle with the GRE’s time constraints, and you probably have difficulties with probability questions.

Great! In this article, we’ll examine how probability questions can provide you with a convenient opportunity to make up lost time.

To set this up, please consider the following scenario:

*It’s test day, and halfway through one of the math sections, you find that you’re 2 minutes behind.*

At this point, you have two options:

1) Work faster on the remaining questions (and risk making careless mistakes)

2) Guess on one of the questions and immediately make up the lost time (but risk guessing the wrong answer)

Both options are less than ideal, but I’ll argue that option #2 is better than option #1, especially if you encounter a probability question.

To illustrate this, answer the following question within 20 seconds:

**From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected?**

**(A) 0.1**

**(B) 0.2**

**(C) 0.25**

**(D) 0.4 **

**(E) 0.6**

If you’ve already identified probability as one of your weaknesses, and if you typically fall behind time-wise, this question is a gift. You should be able to eliminate 2 or 3 answer choices and make an educated guess within seconds of reading the question.

The elimination strategy relies on the fact that most people have an innate ability to judge the relative likelihood of an event. So, for the Las Vegas question above, you can use your intuition to eliminate answer choices that just don’t *feel* right.

To begin, you might ask, “Is the probability of selecting Marnie and Noomi greater than 0.5 or less than 0.5?” If it feels less than 0.5 (which it is), you can eliminate E. Of course, you’ll want to eliminate more than 1 answer choice, so you’ll need to be more aggressive. You might ask, “Does the event seem *very* unlikely or a *little* unlikely? Your answer will allow you to eliminate additional answer choices.

If you feel that the probability seems very unlikely, you might eliminate C, D and E, leaving yourself with a good chance of guessing the correct answer (all within seconds of reading the question). If you’re less aggressive, you might eliminate just D and E. That’s still fine. Remember that the goal here is not to ensure that you correctly answer the question; the goal is to make up your 2 minutes and maximize your chances of guessing the correct answer.

Please note that this guessing strategy can also be used if you typically run out of time on the math sections, and you need a way to give yourself a buffer. Just remember that probability questions are the best for this (counting questions are pretty good, too).

To illustrate how GRE probability questions are better than other question types for guessing, try answering this next question within 20 seconds:

**Gita is 12 years older than Harvey. In 4 years, Gita will be twice as old as Harvey. How old will Harvey be in 3 years? **

**(A) 5**

**(B) 11**

**(C) 15**

**(D) 19 **

**(E) 23**

Although this question a little easier than the Las Vegas question, it’s a bad candidate for guessing (and recapturing lost time), since we really can’t use our intuition to legitimately eliminate answer choices.

In my next article, I’ll talk more about how we can further eliminate answer choices in probability questions before we guess. In that article, I plan to continue discussing the Las Vegas question, so I won’t tell you the answer just yet. I will, however, tell you that the answer to the Gita/Harvey question is B.

Hi Chris,

i was just wondering when they say that you considered that in four years gita will be twice as old as harvey will be in for years not how old he is now. and can you answer the first question.

Hi Rihan 🙂

The phrase “In 4 years, Gita will be twice as old as Harvey” is referring to Gita and Harvey’s ages four years from now. That is to say if

G = Gita’s age now

H = Harvey’s age now

G + 4 = Gita’s age in 4 years

H + 4 = Harvey’s age in 4 years

And we could translate the statement in the prompt as:

G + 4 = 2(H + 4).We also know from the question that in the present, Gita is 12 years older than Harvey: G = H + 12.

So, we have 2 equations and 2 variables:

* G + 4 = 2(H + 4)

* G = H + 12

And in the end, we want to find the value H + 3, so first, let’s use the two equations to find the value of H. We’ll replace G with H + 12 in the first equation:

G + 4 = 2(H + 4) –> H + 12 + 4 = 2(H + 4)

H + 16 = 2H + 8

8 = H

So, in the present Harvey is 8 years old. That means in 3 years from the present, Harvey will be 11 years old 🙂

Let Harvey’s present be x , Then Gita’s age is x+14. In 4 years time, Harvey’s age will be x+4 and Gita’s age will be x+16. Also in 4 years time, Gita’s age = 2 the age of Harvey. Therefore,

x+16 =2(x+4). Solving gives x = 8 years. So in 3 years time Gita’s age will be x+3 =8+3 == 11 years (ANS = C)

Why does x+14 turn into x+16 after 4 years?

Hello sir

For the question

Gita is 12 years older than Harvey. In 4 years, Gita will be twice as old as Harvey. How old will Harvey be in 3 years?

I answer it as bellow /

G/ for old of Gita , H/ for old of Harvey!

G=H+12

G+4=2H

Sub 2 in 1

We get G= 28

H= 16

H after 3 years will be 16+ 3 = 19

Answe is C ,

I hope you can explain to me

Best

Hi Ameer,

The step you missed was in translating the “In 4 years, Gita will be twice as old as Harvey”.

You added 4 to Gita but not to Harvey. Therefore, the equation should read

G+4 = 2(H+4),

G+4 = 2H + 8…by using substitution, you should get to answer (B).

Hope that helps!

HARRY = H

GITA = H+12

AFTER 4 YEARS:

HARRY = (H)+4

GITA = (H+12)+4

Gita will be twice as old as Harvey:

2(H+4)

2H+8 = H+16

2H-H = 16-8

H = 8

So,

H = 8

H+12 = GITA = 8+12 =20

In 4 years, Gita will be twice as old as Harvey:

in 4 years gita (24years) will be twice as old as Harvey (8+4 = 12 years)

How old will Harvey be in 3 years ?

Harvey (12 + 3) = 15 years

CAN YOU PLZ POST THE ANSWER FOR THE QUESTION From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected?

Hi Muhammad,

You can find an explanation for that practice question in this post on our blog. In the post, we provide two different methods to solve the problem 😀

* Rules and Counting Techniques for Probability on the GRE

Hope the post helps 🙂

HARRY = H

GITA = H+12

AFTER 4 YEARS:

HARRY = (H)+4

GITA = (H+12)+4

Gita will be twice as old as Harvey:

2(H+4)

2H+8 = H+16

2H-H = 16-8

H = 8

So,

H = 8 = HARRY

H+12 = GITA = 8+12 =20

In 4 years, Gita will be twice as old as Harvey:

in 4 years gita (24years) will be twice as old as Harvey (8+4 = 12 years)

How old will Harvey be in 3 years ?

Harvey (12 + 3) = 15 years

Let Harvey’s present age be x

Then, Gita’s present age = x + 12

After 4 years Gita’s and Harvey’s age will be x+12+4 and x+4 respectively.

Thus, the equation is as follows:

x + 12 + 4 = 2( x + 4)

x + 16 = 2x + 8

sub -x from both sides,

16 = x + 8

sub -8 from both sides,

x = 8

Hence the present age of Harvey is 8 yrs, and three years from now he will be 11 yrs.

Thus choice B is the correct ans.

NICE

Could you please explain the correct strategy used to find the correct answer to the Las Vegas question? I still cannot figure it out and my math skills are not the best.

Hi shara,

I plan to discuss strategies over the next two posts.

Cheers,

Brent

Update:

Here are the links

http://magoosh.com/gre/2011/gre-probability-questions-%E2%80%93-rules-and-counting-techniques/

http://magoosh.com/gre/2011/gre-probability-questions-using-the-elimination-strategy/