From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

(A) 3/7
(B) 5/12
(C) 27/70
(D) 2/7
(E) 9/35

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event

A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: Blue, Green, Yellow or Pink. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color or 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible?

(A) 6
(B) 8
(C) 16
(D) 24
(E) 32

A certain office supply store stocks 2 sizes of self-stick notepads each in 4 colors Blue Green Yellow or Pink

 

A basket contains 5 apples, of which 1 is spoiled and the rest are good. If Henry is to select 2 apples from the basket simultaneously and at random, what is the possibility that the 2 apples selected will include the spoiled apple?

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

GMAT tutoring students tend to dislike probability questions. A couple of things to remember regarding GMAT probability:

  1. It doesn't come up that much.
  2. When it does come up often the question isn't that challenging.
  3. If you do get a challenging probability question you can always skip.

Let's focus on the question first: what is the possibility that the 2 apples selected will include the spoiled apple?

We need a spoiled apple selected. OK. There's a very practical way of doing this:

G1 G2 G3 G4 S

10 Possible pairs. 4 of them have a spoiled apple. So 4/10 or 2/5 chance of having a pair with the spoiled apple.

G1 G2

G1 G3

G1 G4

G1 S

G2 G3

G2 G4

G S

G3 G4

G3 S

G4 S

Easy. If the numbers are small there's nothing wrong with writing things out. In fact, it can be the best approach.

The other way to do it is with the slot method. Calculate the number of ways to create a group of two from 5 things. That will be your total or your denominator. And then calculate the number ways to create a group of two with the constraint that one of those things (the spoiled apple) must be included. That will be your numerator. Remember that probability is just specific scenario/all scenarios.

5*4/1*2 = 10 (the number of ways to create a pair from 5 things)

With one of the spots reserved for the spoiled apple you're only left with the other spot to populate. So how many things can go in that spot? 4 (the 4 good apples). So the numerator is 4 and the denominator is 10. 4/10 = 2/5.

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 - by^2, where b is an integer?

(A) 1/2

(B) 1/3

(C) 1/4

(D) 1/5

(E) 1/6

Be super comfortable with difference of squares. You're almost guaranteed to need it on your GMAT. You need to know it forwards, backwards, and upside-down. There isn't much content you need to know for the GMAT quant but in order to be consistently successful so that you can go in there on test day and snag your 700+ score you need complete fluency with that limited content. On this one we're looking for a product of two terms that will produce difference of squares. So we need something in the format: (x+y(x-y). There's only one pairing that gives us the right format. Now you have to figure out what that means in terms of probability. In general, probability is specific scenario/total scenarios. You can calculate the total and the specific scenarios using the slot method. How many teams of 2 can you make from four things? (4*3)/(1*2) = 6. So that's your denominator. And then there's only one team that works, the pairing of (x + y)(x - y). So that leaves you with 1/6. E. Detailed diagrams below and an in depth video explantation here: If two of the four expressions

 

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random

You also can think about this in terms of straight probability and say: 2 choices our of 4 and then 1 choice out of 3. 2/4 * 1/3 = 2/12 = 1/6. Comment with any questions or additions. Happy studies!

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random alt

 

 

GMAT Question of the Day - Data Sufficiency - Combinatorics

A team is composed of exactly 4 executives and 3 associates. How many such teams can be formed?

(1) If two fewer executives were available for selection there would be equal an number of executives and associates available for selection.

(2) If two more associated were available for selection exactly 10 different teams of 3 associates could be selected.

Answer Show

GMAT Question of the Day Solution

GMAT combinatorics questions do not have to be difficult. They seem challenging because they are opaque. It's not always easy to see through them. Often though, there is very little that you need to do in order to get the answer. As with the rest of the GMAT it's all about getting the right context. Categorizing the question. Are you putting together a team or creating an order? If it is an order is it unique? How many "players" do you have? How many slots on the team do you have for these players? In this GMAT question of the day we have 7 slots which are separated into executive slots and associate slots. We're not given the number of executives/associates. We'll need to figure that out with the statements (if possible).

Statement 1 tells us that that executives = associates + 2 but does not tell us anything about the real numbers. So there could be 998 associates and 1000 executives or 5 associates and 7 executives. Those two sets of values would produce different numbers of teams. Insufficient.

Statement 2 tells us that we have 3 associates but we have no information on the executives.

Putting the together we have the number of people in each group and can calculate the number of teams. Hurray.

GMAT Question of the Day Data Sufficiency Combinatorics Solution Diagram

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