GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 20 Jun 2018, 15:54

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Math: Probability

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
162 KUDOS received
CEO
CEO
User avatar
B
Joined: 17 Nov 2007
Posts: 3486
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Math: Probability [#permalink]

Show Tags

New post Updated on: 26 Apr 2018, 04:25
162
176
PROBABILITY

Image
This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

Image
--------------------------------------------------------
Get The Official GMAT Club's App - GMAT TOOLKIT 2.
The only app you need to get 700+ score!

[iOS App] [Android App]

--------------------------------------------------------

Definition

A number expressing the probability (p) that a specific event will occur, expressed as the ratio of the number of actual occurrences (n) to the number of possible occurrences (N).

\(p = \frac{n}{N}\)

A number expressing the probability (q) that a specific event will not occur:

\(q = \frac{(N-n)}{N} = 1 - p\)

Examples

Coin

ImageImage

There are two equally possible outcomes when we toss a coin: a head (H) or tail (T). Therefore, the probability of getting head is 50% or \(\frac{1}{2}\) and the probability of getting tail is 50% or \(\frac{1}{2}\).
All possibilities: {H,T}

Dice

Image

There are 6 equally possible outcomes when we roll a die. The probability of getting any number out of 1-6 is \(\frac{1}{6}\).
All possibilities: {1,2,3,4,5,6}

Marbles, Balls, Cards...

Image

Let's assume we have a jar with 10 green and 90 white marbles. If we randomly choose a marble, what is the probability of getting a green marble?
The number of all marbles: N = 10 + 90 =100
The number of green marbles: n = 10
Probability of getting a green marble: \(p = \frac{n}{N} = \frac{10}{100} = \frac{1}{10}\)

There is one important concept in problems with marbles/cards/balls. When the first marble is removed from a jar and not replaced, the probability for the second marble differs (\(\frac{9}{99}\) vs. \(\frac{10}{100}\)). Whereas in case of a coin or dice the probabilities are always the same (\(\frac{1}{6}\) and \(\frac{1}{2}\)). Usually, a problem explicitly states: it is a problem with replacement or without replacement.


Independent events

Two events are independent if occurrence of one event does not influence occurrence of other events. For n independent events the probability is the product of all probabilities of independent events:

p = p1 * p2 * ... * pn-1 * pn

or

P(A and B) = P(A) * P(B) - A and B denote independent events

Example #1
Q:There is a coin and a die. After one flip and one toss, what is the probability of getting heads and a "4"?
Solution: Tossing a coin and rolling a die are independent events. The probability of getting heads is \(\frac{1}{2}\) and probability of getting a "4" is \(\frac{1}{6}\). Therefore, the probability of getting heads and a "4" is:
\(P = \frac{1}{2} * \frac{1}{6} = \frac{1}{12}\)

Example #2
Q: If there is a 20% chance of rain, what is the probability that it will rain on the first day but not on the second?
Solution: The probability of rain is 0.2; therefore probability of sunshine is q = 1 - 0.2 = 0.8. This yields that the probability of rain on the first day and sunshine on the second day is:
P = 0.2 * 0.8 = 0.16

Example #3
Q:There are two sets of integers: {1,3,6,7,8} and {3,5,2}. If Robert chooses randomly one integer from the first set and one integer from the second set, what is the probability of getting two odd integers?
Solution: There is a total of 5 integers in the first set and 3 of them are odd: {1, 3, 7}. Therefore, the probability of getting odd integer out of first set is \(\frac{3}{5}\). There are 3 integers in the second set and 2 of them are odd: {3, 5}. Therefore, the probability of getting an odd integer out of second set is \(\frac{2}{3}\). Finally, the probability of of getting two odd integers is:
\(P = \frac{3}{5} * \frac{2}{3} = \frac{2}{5}\)


Mutually exclusive events

Shakespeare's phrase "To be, or not to be: that is the question" is an example of two mutually exclusive events.

Two events are mutually exclusive if they cannot occur at the same time. For n mutually exclusive events the probability is the sum of all probabilities of events:

p = p1 + p2 + ... + pn-1 + pn

or

P(A or B) = P(A) + P(B) - A and B denotes mutually exclusive events

Example #1
Q: If Jessica rolls a die, what is the probability of getting at least a "3"?
Solution: There are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The probability of each outcome is 1/6. The probability of getting at least a "3" is:
\(P = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{2}{3}\)


Combination of independent and mutually exclusive events

Many probability problems contain combination of both independent and mutually exclusive events. To solve those problems it is important to identify all events and their types. One of the typical problems can be presented in a following general form:

Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence?
(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)
Solution: All events are independent. So, we can say that:

\(P' = p^k*(1-p)^{n-k}\) (1)

But it isn't the right answer. It would be right if we specified exactly each position for events in the sequence. So, we need to take into account that there are more than one outcomes. Let's consider our example with a coin where "H" stands for Heads and "T" stands for Tails:
HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3 heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In our general question, probability of occurring event k times in n-time sequence could be expressed as:

\(P = C^n_k*p^k*(1-p)^{n-k}\) (2)

In the example with a coin, right answer is \(P = C^8_3*0.5^3*0.5^5 =C^8_3*0.5^8\)

Example #1
Q.:If the probability of raining on any given day in Atlanta is 40 percent, what is the probability of raining on exactly 2 days in a 7-day period?
Solution: We are not interested in the exact sequence of event and thus apply formula #2:
\(P = C^7_2*0.4^2*0.6^5\)


A few ways to approach a probability problem

There are a few typical ways that you can use for solving probability questions. Let's consider example, how it is possible to apply different approaches:

Example #1
Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?
Solution:

1) combinatorial approach: The total number of possible committees is \(N=C^8_2\). The number of possible committee that includes both Bob and Rachel is \(n=1\).
\(P = \frac{n}{N} = \frac{1}{C^8_2} = \frac{1}{28}\)

2) reversal combinatorial approach: Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula p = 1 - q. The total number of possible committees is \(N=C^8_2\). The number of possible committee that does not includes both Bob and Rachel is:
\(m = C^6_2 + 2*C^6_1\) where,
\(C^6_2\) - the number of committees formed from 6 other people.
\(2*C^6_1\) - the number of committees formed from Rob or Rachel and one out of 6 other people.
\(P = 1- \frac{m}{N} = 1 - \frac{C^6_2 + 2*C^6_1}{C^8_2}\)
\(P = 1 - \frac{15+2*6}{28} = 1 - \frac{27}{28} = \frac{1}{28}\)

3) probability approach: The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is.
\(P = \frac{2}{8} * \frac{1}{7} = \frac{2}{56} = \frac{1}{28}\)

4) reversal probability approach: We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is.
\(P = 1 - (\frac{2}{8} * \frac{6}{7} + \frac{6}{8} * 1) = \frac{2}{56} = \frac{1}{28}\)

Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?
Solution:

1) combinatorial approach:
\(C^5_3\) - we choose 3 couples out of 5 couples.
\(C^2_1\) - we chose one person out of a couple.
\((C^2_1)^3\) - we have 3 couple and we choose one person out of each couple.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=\frac{C^5_3*(C^2_1)^3}{C^{10}_3}=\frac{10*8}{10*3*4} = \frac{2}{3}\)

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

3) probability approach:
1st person: \(\frac{10}{10} = 1\) - we choose any person out of 10.
2nd person: \(\frac{8}{9}\) - we choose any person out of 8=10-2(one couple from previous choice)
3rd person: \(\frac{6}{8}\) - we choose any person out of 6=10-4(two couples from previous choices).

\(p = 1*\frac{8}{9}*\frac{6}{8}=\frac{2}{3}\)


Probability tree

Sometimes, at 700+ level you may see complex probability problems that include conditions or restrictions. For such problems it could be helpful to draw a probability tree that include all possible outcomes and their probabilities.

Example #1
Q: Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?
Solution: Let's draw all possible outcomes and calculate all probabilities.

Image

Now, It is pretty obvious that the probability of Julia's win is:
\(P = \frac47 + \frac27*\frac16 + \frac17*\frac26 = \frac23\)


Tips and Tricks: Symmetry

Symmetry sometimes lets you solve seemingly complex probability problem in a few seconds. Let's consider an example:

Example #1
Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ?
Solution: Because of symmetry, the number of ways that Bob is left to Rachel is exactly 1/2 of all possible ways:
\(N = \frac12*P^5_2 = 10\)


Official GMAC Books:

The Official Guide, 12th Edition: DT #4; DT #7; PS #12; PS #67; PS #105; PS #158; PS #174; PS #214; DS #3; DS #107;
The Official Guide, Quantitative 2th Edition: PS #79; PS #160;
The Official Guide, 11th Edition: DT #4; DT #7; PS #10; PS #64; PS #173; PS #217; PS #231; DS #82; DS #114;

Generated from [GMAT ToolKit 2]


Resources

Probability DS problems: [search]
Probability PS problems: [search]

Walker's post with Combinatorics/probability problems: [Combinatorics/probability Problems]
Bullet's post with probability problems: [Combined Probability Questions]

Image
--------------------------------------------------------
Get The Official GMAT Club's App - GMAT TOOLKIT 2.
The only app you need to get 700+ score!

[iOS App] [Android App]

--------------------------------------------------------


Spoiler: :: Images
Attachment:
Math_probability_I_head.png
Math_probability_I_head.png [ 31.29 KiB | Viewed 187707 times ]
Attachment:
Math_probability_I_tail.png
Math_probability_I_tail.png [ 32.04 KiB | Viewed 187492 times ]
Attachment:
Math_probability_I_dice.png
Math_probability_I_dice.png [ 12.38 KiB | Viewed 187267 times ]
Attachment:
Math_probability_I_marbles.png
Math_probability_I_marbles.png [ 31.81 KiB | Viewed 187156 times ]
Attachment:
Math_probability_tree.png
Math_probability_tree.png [ 9.84 KiB | Viewed 186857 times ]
Attachment:
Math_icon_probability.png
Math_icon_probability.png [ 4.65 KiB | Viewed 182559 times ]

_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame


Originally posted by walker on 23 Nov 2009, 22:17.
Last edited by Bunuel on 26 Apr 2018, 04:25, edited 26 times in total.
Edited.
8 KUDOS received
Intern
Intern
User avatar
Joined: 19 Feb 2009
Posts: 47
Schools: INSEAD,Nanyang Business school, CBS,
Re: Math: Probability [#permalink]

Show Tags

New post 25 Nov 2009, 05:02
8
Man....it must have taken lots of efforts to compile this document.....

kudos to u...


Thanx for sharing this...
_________________

Working without expecting fruit helps in mastering the art of doing fault-free action !

Expert Post
6 KUDOS received
Founder
Founder
User avatar
V
Joined: 04 Dec 2002
Posts: 16980
Location: United States (WA)
GMAT 1: 750 Q49 V42
GMAT ToolKit User CAT Tests
Re: Math: Probability [#permalink]

Show Tags

New post 27 Nov 2009, 17:19
6
Wow - between Walker and Bunuel, we have Math covered.
Great job!
_________________

Founder of GMAT Club

Just starting out with GMAT? Start here... or use our Daily Study Plan


Co-author of the GMAT Club tests

4 KUDOS received
Intern
Intern
avatar
Joined: 24 Oct 2009
Posts: 5
Re: Math: Probability [#permalink]

Show Tags

New post 03 Dec 2009, 17:17
4
3
Example #1
Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ?

Slot method to solve this problem:

Chair location: 1 2 3 4 5


R has 4 possible choices (2, 3, 4, 5).

When R is on chair 5, B can have 4 options to choose from (1, 2, 3, 4) that are left of chair 5 = 4
When R is on chair 4, B can have 3 options to choose from (1, 2, 3) that are left of chair 4 = 3
When R is on chair 3, B can have 2 options to choose from (1, 2) that are left of chair 3 = 2
When R is on chair 2, B can have 1 option to choose from (1) that is left of chair 2 = 1

Possible ways = 4 + 3 + 2 + 1 = 10
Intern
Intern
User avatar
Joined: 21 Feb 2010
Posts: 26
Location: Ukraine
Re: Math: Probability [#permalink]

Show Tags

New post 17 Mar 2010, 12:19
Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence?
(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)
Solution: All events are independent. So, we can say that:

\(P' = p^k*(1-p)^{n-k}\) (1)

But it isn't the right answer. It would be right if we specified exactly each position for events in the sequence. So, we need to take into account that there are more than one outcomes. Let's consider our example with a coin where "H" stands for Heads and "T" stands for Tails:
HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3 heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In our general question, probability of occurring event k times in n-time sequence could be expressed as:

\(P = C^n_k*p^k*(1-p)^{n-k}\) (2)

In the example with a coin, right answer is \(P = C^8_3*0.5^3*0.5^5 =C^8_3*0.5^8\)

I didn't get how we have found 0.5. The power 3 - is it mean the number of needed H and the power 5 represent T? So, what is 0.5?

Example #1
Q.:If the probability of raining on any given day in Atlanta is 40 percent, what is the probability of raining on exactly 2 days in a 7-day period?
Solution: We are not interested in the exact sequence of event and thus apply formula #2:
\(P = C^7_2*0.4^2*0.6^5\)


I want to specify, the powers 2 and 5 are the numbers of days?? I assume that 0,4 it's the likelihood of the raining and 0.6 of the sunny weather. Right?
Expert Post
CEO
CEO
User avatar
B
Joined: 17 Nov 2007
Posts: 3486
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: Math: Probability [#permalink]

Show Tags

New post 18 Mar 2010, 23:06
fruit wrote:
I didn't get how we have found 0.5. The power 3 - is it mean the number of needed H and the power 5 represent T? So, what is 0.5?

0.5 is probability of getting head (or tail).

fruit wrote:

I want to specify, the powers 2 and 5 are the numbers of days?? I assume that 0,4 it's the likelihood of the raining and 0.6 of the sunny weather. Right?

That's right.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Manager
User avatar
Joined: 30 Nov 2010
Posts: 230
Schools: UC Berkley, UCLA
Re: Math: Probability [#permalink]

Show Tags

New post 15 Dec 2010, 08:15
Example #1
Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?
Solution:

1) combinatorial approach: The total number of possible committees is N=C^8_2. The number of possible committee that includes both Bob and Rachel is n=1.
P = \frac{n}{N} = \frac{1}{C^8_2} = \frac{1}{28}

2) reversal combinatorial approach: Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula p = 1 - q. The total number of possible committees is N=C^8_2. The number of possible committee that does not includes both Bob and Rachel is:
m = C^6_2 + 2*C^6_1 where,
C^6_2 - the number of committees formed from 6 other people.
2*C^6_1 - the number of committees formed from Rob or Rachel and one out of 6 other people. Why is it multiplied by 2? Can someone pls explain it to me?
P = 1- \frac{m}{N} = 1 - \frac{C^6_2 + 2*C^6_1}{C^8_2}
P = 1 - \frac{15+2*6}{28} = 1 - \frac{27}{28} = \frac{1}{28}
_________________

Thank you for your kudoses Everyone!!!


"It always seems impossible until its done."
-Nelson Mandela

Expert Post
1 KUDOS received
CEO
CEO
User avatar
B
Joined: 17 Nov 2007
Posts: 3486
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User Premium Member
Re: Math: Probability [#permalink]

Show Tags

New post 15 Dec 2010, 08:33
1
\(C^6_1\) - the number of committees formed from Bob and one person out of 6 people.
\(C^6_1\) - the number of committees formed from Rachel and one person out of 6 people.

So, \(C^6_1 + C^6_1 = 2*C^6_1\) - the number of committees formed from Rob or Rachel and one person out of 6 other people.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Intern
avatar
Joined: 28 Dec 2010
Posts: 4
Re: Math: Probability [#permalink]

Show Tags

New post 04 Mar 2011, 15:36
I'm having a little trouble with the solution to a sample problem in MGMAT Word Translations (3rd Ed) p. 99-100.

Q: "A medical researcher must choose one of 14 patients to receive an experimental medicine called Progaine. The researcher must then choose one of the remaining 13 patients to receiver another medicine, called Ropecia. Finally, the researcher administers a placebo to one of the remaining 12 patients. All choices are equally random. If Donald is one of the 14 patients, what is the probability that Donald receivers either Prograine or Ropecia?"

I thought this was a simple "without replacement" question.

Probability of Donald getting Progaine: 1/14
Probability of Donald getting Ropecia after one patient is removed from the pool: 1/13

(1/14)+(1/13) = 27/182

The book says the answer is

Probability of Donald getting Prograine: 1/14
Probability of Donald getting Ropecia: 1/14

(1/14)+(1/14) = 1/7

Anyone see why this is not a "without replacement" problem? I'm stumped.
2 KUDOS received
Math Forum Moderator
avatar
Joined: 20 Dec 2010
Posts: 1900
Re: Math: Probability [#permalink]

Show Tags

New post 04 Mar 2011, 16:03
2
1
jasonedward wrote:
I'm having a little trouble with the solution to a sample problem in MGMAT Word Translations (3rd Ed) p. 99-100.

Q: "A medical researcher must choose one of 14 patients to receive an experimental medicine called Progaine. The researcher must then choose one of the remaining 13 patients to receiver another medicine, called Ropecia. Finally, the researcher administers a placebo to one of the remaining 12 patients. All choices are equally random. If Donald is one of the 14 patients, what is the probability that Donald receivers either Prograine or Ropecia?"

I thought this was a simple "without replacement" question.

Probability of Donald getting Progaine: 1/14
Probability of Donald getting Ropecia after one patient is removed from the pool: 1/13-- something wrong here

(1/14)+(1/13) = 27/182

The book says the answer is

Probability of Donald getting Prograine: 1/14
Probability of Donald getting Ropecia: 1/14

(1/14)+(1/14) = 1/7

Anyone see why this is not a "without replacement" problem? I'm stumped.


Donald Getting prograine = 1/14 (1 out of 14 patients should be chosen for prograine)
Donald Not getting prograine = 13/14
Donald Getting Ropecia = 1/13 (1 out of 13 patients should be chosen for Ropecia)

How can Donald get either P or R
Donald will get Prograine and get done OR Donald will not get Prograine and get Ropecia

1/14+ 13/14*1/13 = 1/14+1/14 = 1/7

I believe it is a without replacement problem.
_________________

~fluke

GMAT Club Premium Membership - big benefits and savings

Expert Post
6 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46207
Re: Math: Probability [#permalink]

Show Tags

New post 05 Mar 2011, 03:46
6
jasonedward wrote:
I'm having a little trouble with the solution to a sample problem in MGMAT Word Translations (3rd Ed) p. 99-100.

Q: "A medical researcher must choose one of 14 patients to receive an experimental medicine called Progaine. The researcher must then choose one of the remaining 13 patients to receiver another medicine, called Ropecia. Finally, the researcher administers a placebo to one of the remaining 12 patients. All choices are equally random. If Donald is one of the 14 patients, what is the probability that Donald receivers either Prograine or Ropecia?"

I thought this was a simple "without replacement" question.

Probability of Donald getting Progaine: 1/14
Probability of Donald getting Ropecia after one patient is removed from the pool: 1/13

(1/14)+(1/13) = 27/182

The book says the answer is

Probability of Donald getting Prograine: 1/14
Probability of Donald getting Ropecia: 1/14

(1/14)+(1/14) = 1/7

Anyone see why this is not a "without replacement" problem? I'm stumped.


Donald to receiver either Prograine or Ropecia must be among first two chosen patients and as there are 14 patients then the probability of this is simply 2/14=1/7.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

1 KUDOS received
Intern
Intern
avatar
Joined: 24 Apr 2011
Posts: 11
Location: India
Schools: CBS, Booth, UNC,ISB
WE 1: Software Develoment
WE 2: Derivative Trader
Re: Math: Probability [#permalink]

Show Tags

New post 20 May 2011, 00:37
1
1
Let me briefly talk about a concept that i remember studying long back.Most of the logic would be excerpts from Higher Algebra by Bernard Shaw

We know that most of probability questions can be easily cracked if we have a strong grip on Permutation and combination.Let me present you all with something interesting. I'll be dealing with "Distribution" of similar and dissimilar things

Concept 1: Distributing Similar things among people where each recipient receives at least 1 thing
----------------------------------------------------------------------------------------------

Imagine you have 10 similar tennis balls which you need distribute it to 3 friends and each of the fella receives at least 1 ball.

how would you approach it???

Let us do it figuratively.Imagine the 10 balls

O O O O O O O O O O ( remember there are 9 spaces between 10 balls)

if you consider 2 sticks ! !, how many ways can u place them in the 9 spaces between the 10 balls. yeah right 9C2 ways.. would u believe , thats the answer. for example consider O O ! O O O ! O O O O O . this is one of the 9C2 ways where the first guy gets 2 balls, 2 nd guy gets 3 and the last gets 5. you just need to place (r-1) sticks in the spaces between the n balls and that's like selecting r-1 spaces between the n balls.
so the answer is n-1Cr-1 n= number of balls r= number of recepients
I can rephrase the same question

Find the number of solution to the equation
X + Y + Z = 10 where each X, Y and Z integers and are >0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls


hence u already know the solution 9C 2 :shock:


Concept 2 : Distributing Similar things among people where each recipient may receive no balls at all
----------------------------------------------------------------------------------------------

Imagine the same old story where you have 10 similar tennis balls which you need distribute it to 5 friends. However now 1 or more of them may receive 0 balls

Again imagine the balls lying beautifully in the imagination of your brain

O O O O O O O O O O .

Now as you know for 5 friends you need to consider 4 sticks ! ! ! !. Now that any of the friend may receive 0 balls also, we cannot consider the space between the balls. So what do you think.. let me figuratively sample u some solution

one of the solution O ! O ! O O ! O O O ! O O O ( FRIEND 1= 1 BALL, FRIEND 2= 1 BALL FRIEND 3= 2 BALLS FRIEND 4= 3 BALLS FRIEND 5 = 3BALLS)
ANOTHER SOLUTION ! O O O ! O O O ! O O O O ! ( FRIEND 1 = 0 BALLS , FRIEND 2= 3 BALLS , FRIEND 3= 3 BALLS , FRIEND 4 = 4 BALLS FRIEND 5 = 0)

I hope by now you would have guessed the solution is nothing but arranging 14 things of which 10 are of one kind and 4 are of one kind

that is !14/(!10*!4) . this is nothing but n+r-1 C n where n is the number of balls and r is the number of recipients

I can again rephrase the same question

Find the number of solution to the equation
X + Y + Z = 10 where each X, Y and Z integers and are >=0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls


hence u already know the solution 14C4 :shock:



I can present some more topics. this will be based on the condition that the readers find it interesting and helpful
Manager
Manager
avatar
Joined: 30 May 2008
Posts: 63
Re: Math: Probability [#permalink]

Show Tags

New post 11 Apr 2012, 23:36
walker wrote:
PROBABILITY

Image
This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

--------------------------------------------------------

Image

This topic is included in GMAT ToolKit App (iPhone/iPod Touch)
[read more] [AppStore]



--------------------------------------------------------

A few ways to approach a probability problem

There are a few typical ways that you can use for solving probability questions. Let's consider example, how it is possible to apply different approaches:

Example #1
Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?
Solution:

3) probability approach: The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is.
\(P = \frac{2}{8} * \frac{1}{7} = \frac{2}{56} = \frac{1}{28}\)

4) reversal probability approach: We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is.
\(P = 1 - (\frac{2}{8} * \frac{6}{7} + \frac{6}{8} * 1) = \frac{2}{56} = \frac{1}{28}\)


Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?
Solution:

1) combinatorial approach:
\(C^5_3\) - we choose 3 couples out of 5 couples.
\(C^2_1\) - we chose one person out of a couple.
\((C^2_1)^3\) - we have 3 couple and we choose one person out of each couple.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=\frac{C^5_3*(C^2_1)^3}{C^{10}_3}=\frac{10*8}{10*3*4} = \frac{2}{3}\)

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

3) probability approach:
1st person: \(\frac{10}{10} = 1\) - we choose any person out of 10.
2nd person: \(\frac{8}{9}\) - we choose any person out of 8=10-2(one couple from previous choice)
3rd person: \(\frac{6}{8}\) - we choose any person out of 6=10-4(two couples from previous choices).

\(p = 1*\frac{8}{9}*\frac{6}{8}=\frac{2}{3}\)


Probability tree


Official GMAC Books:

The Official Guide, 12th Edition: DT #4; DT #7; PS #12; PS #67; PS #105; PS #158; PS #174; PS #214; DS #3; DS #107;
The Official Guide, Quantitative 2th Edition: PS #79; PS #160;
The Official Guide, 11th Edition: DT #4; DT #7; PS #10; PS #64; PS #173; PS #217; PS #231; DS #82; DS #114;

Generated from [GMAT ToolKit]


Resources

Probability DS problems: [search]
Probability PS problems: [search]

Walker's post with Combinatorics/probability problems: [Combinatorics/probability Problems]
Bullet's post with probability problems: [Combined Probability Questions]
-------------------------------------------------------

Spoiler: :: Images
Attachment:
Math_probability_I_head.png
Attachment:
Math_probability_I_tail.png
Attachment:
Math_probability_I_dice.png
Attachment:
Math_probability_I_marbles.png
Attachment:
Math_probability_tree.png
Attachment:
Math_icon_probability.png



Can someone explain the reversal probability approach under the first bold face part?

Also please please explain the red text under the married couple example, how did you know to choose 3 couple out of 5? Is it because the question asks for 3 ppl not married? Im completely lost on this example and particularly on the reversal combinatorial approach......
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46207
Re: Math: Probability [#permalink]

Show Tags

New post 12 Apr 2012, 00:48
catty2004 wrote:
Can someone explain the reversal probability approach under the first bold face part?

Also please please explain the red text under the married couple example, how did you know to choose 3 couple out of 5? Is it because the question asks for 3 ppl not married? Im completely lost on this example and particularly on the reversal combinatorial approach......


Check similar problems with different approaches:
combination-permutation-problem-couples-98533.html
ps-combinations-94068.html
committee-of-88772.html

Hope it helps.

As for the reverse probability approach, can you please be a bit more specific, what part is confusing for you?
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 21 Mar 2013
Posts: 39
GMAT Date: 03-20-2014
GMAT ToolKit User
Re: Math: Probability [#permalink]

Show Tags

New post 26 Dec 2013, 06:09
fruit wrote:
Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence?
(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)

Are there any GMAT questions testing this concept?
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46207
Re: Math: Probability [#permalink]

Show Tags

New post 26 Dec 2013, 06:29
idinuv wrote:
fruit wrote:
Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence?
(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)

Are there any GMAT questions testing this concept?


Questions on this subject:
there-is-a-90-chance-that-a-registered-voter-in-burghtown-56812.html
a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html
the-probability-that-a-family-with-6-children-has-exactly-88945.html
kate-and-david-each-have-10-together-they-flip-a-coin-97177.html
a-fair-2-sided-coin-is-flipped-6-times-what-is-the-88069.html
how-many-different-6-letter-sequences-are-there-that-consist-96468.html
a-fair-2-sided-coin-is-flipped-6-times-what-is-the-87673.html
every-trading-day-the-price-of-cf-corp-stock-either-goes-up-98093.html
what-is-the-probability-that-a-field-gun-will-hit-thrice-on-127334.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html


Hope this helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
User avatar
Joined: 18 Sep 2013
Posts: 7
Re: Math: Probability [#permalink]

Show Tags

New post 12 Mar 2014, 11:22
Dear Walker

I was trying your given example- Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other? with other method but the answer is coming wrong.

I did 10c1X8c1X6c1/ 10c3. We can select any one random person from 10 then to avoid pairing we remove his/her counterpart so next selection becomes 8c1 and similarly 6c1...but the probability is coming out to be 4. Where am I wrong?

Thanks for your help! :)
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46207
Re: Math: Probability [#permalink]

Show Tags

New post 13 Mar 2014, 02:00
ShantnuMathuria wrote:
Dear Walker

I was trying your given example- Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other? with other method but the answer is coming wrong.

I did 10c1X8c1X6c1/ 10c3. We can select any one random person from 10 then to avoid pairing we remove his/her counterpart so next selection becomes 8c1 and similarly 6c1...but the probability is coming out to be 4. Where am I wrong?

Thanks for your help! :)


The following post might help: if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html#p764040

Also check similar questions to practice:
a-committee-of-three-people-is-to-be-chosen-from-four-teams-130617.html
if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html
a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html
a-comittee-of-three-people-is-to-be-chosen-from-four-married-130475.html
a-committee-of-three-people-is-to-be-chosen-from-4-married-101784.html
a-group-of-10-people-consists-of-3-married-couples-and-113785.html
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
given-that-there-are-6-married-couples-if-we-select-only-58640.html
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
B
Joined: 04 Feb 2017
Posts: 9
Re: Math: Probability [#permalink]

Show Tags

New post 25 Feb 2017, 23:21
Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?
Solution:

1) combinatorial approach:
\(C^5_3\) - we choose 3 couples out of 5 couples.
\(C^2_1\) - we chose one person out of a couple.
\((C^2_1)^3\) - we have 3 couple and we choose one person out of each couple.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=\frac{C^5_3*(C^2_1)^3}{C^{10}_3}=\frac{10*8}{10*3*4} = \frac{2}{3}\)

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

3) probability approach:
1st person: \(\frac{10}{10} = 1\) - we choose any person out of 10.
2nd person: \(\frac{8}{9}\) - we choose any person out of 8=10-2(one couple from previous choice)
3rd person: \(\frac{6}{8}\) - we choose any person out of 6=10-4(two couples from previous choices).

\(p = 1*\frac{8}{9}*\frac{6}{8}=\frac{2}{3}\)


--------------------------------------------------------


Didn't follow the reverse combinatorial approach here. The question asks the probability that none of the 3 people selected would be married. But the approach is taking 1 married couple and 1 unmarried person. What am I missing?
1 KUDOS received
Manager
Manager
avatar
D
Joined: 17 May 2015
Posts: 232
CAT Tests
Re: Math: Probability [#permalink]

Show Tags

New post 26 Feb 2017, 05:51
1
Aditi10 wrote:
Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?
Solution:

1) combinatorial approach:
\(C^5_3\) - we choose 3 couples out of 5 couples.
\(C^2_1\) - we chose one person out of a couple.
\((C^2_1)^3\) - we have 3 couple and we choose one person out of each couple.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=\frac{C^5_3*(C^2_1)^3}{C^{10}_3}=\frac{10*8}{10*3*4} = \frac{2}{3}\)

2) reversal combinatorial approach: In this example reversal approach is a bit shorter and faster.
\(C^5_1\) - we choose 1 couple out of 5 couples.
\(C^8_1\) - we chose one person out of remaining 8 people.
\(C^{10}_3\) - the total number of combinations to choose 3 people out of 10 people.

\(p=1 - \frac{C^5_1*C^8_1}{C^{10}_3}=1 - \frac{5*8}{10*3*4} = \frac{2}{3}\)

--------------------------------------------------------


Didn't follow the reverse combinatorial approach here. The question asks the probability that none of the 3 people selected would be married. But the approach is taking 1 married couple and 1 unmarried person. What am I missing?


Aditi10,

We can select 3 people out of 10 peoples (5 married couples) in two possible ways:

1. Event A: None of them are married to each other
or
2. Event B: 1 married couple and 1 person.

Note that event A and B are mutually exclusive and exhaustive events.

We know that total probability is always 1. => P(A) + P(B) = 1

In the first approach, we directly compute P(A).

In "reverse combinatorial" approach, we calculate first P(B) and then subtract it from 1 to get P(A).

Hope it helps.
Re: Math: Probability   [#permalink] 26 Feb 2017, 05:51

Go to page    1   2    Next  [ 28 posts ] 

Display posts from previous: Sort by

Math: Probability

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.