Last visit was: 23 Jul 2024, 04:54 It is currently 23 Jul 2024, 04:54
Toolkit
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

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.

# Math: Probability

SORT BY:
Tags:
Show Tags
Hide Tags
SVP
Joined: 17 Nov 2007
Posts: 2404
Own Kudos [?]: 10144 [664]
Given Kudos: 361
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Intern
Joined: 24 Oct 2009
Posts: 1
Own Kudos [?]: 39 [35]
Given Kudos: 0
Intern
Joined: 24 Apr 2011
Posts: 11
Own Kudos [?]: 23 [23]
Given Kudos: 0
Location: India
Concentration: Finance
Schools:CBS, Booth, UNC,ISB
Q49  V35
GPA: 3.3
WE 1: Software Develoment
General Discussion
Intern
Joined: 19 Feb 2009
Posts: 32
Own Kudos [?]: 593 [19]
Given Kudos: 8
18
Kudos
1
Bookmarks
Man....it must have taken lots of efforts to compile this document.....

kudos to u...

Thanx for sharing this...
Founder
Joined: 04 Dec 2002
Posts: 37864
Own Kudos [?]: 74259 [16]
Given Kudos: 20580
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3
16
Kudos
Wow - between Walker and Bunuel, we have Math covered.
Great job!
Intern
Joined: 21 Feb 2010
Posts: 14
Own Kudos [?]: 15 [1]
Given Kudos: 9
Location: Ukraine
1
Kudos
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?
SVP
Joined: 17 Nov 2007
Posts: 2404
Own Kudos [?]: 10144 [0]
Given Kudos: 361
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
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.
Manager
Joined: 30 Nov 2010
Posts: 189
Own Kudos [?]: 267 [0]
Given Kudos: 66
Schools:UC Berkley, UCLA
GMAT 2: 540 GMAT 3: 530
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}
SVP
Joined: 17 Nov 2007
Posts: 2404
Own Kudos [?]: 10144 [2]
Given Kudos: 361
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
2
Kudos
$$C^6_1$$ - 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.
Intern
Joined: 28 Dec 2010
Posts: 4
Own Kudos [?]: 2 [2]
Given Kudos: 2
1
Kudos
1
Bookmarks
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.
Retired Moderator
Joined: 20 Dec 2010
Posts: 1107
Own Kudos [?]: 4754 [5]
Given Kudos: 376
2
Kudos
3
Bookmarks
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.
Math Expert
Joined: 02 Sep 2009
Posts: 94580
Own Kudos [?]: 643201 [6]
Given Kudos: 86728
6
Kudos
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.
Intern
Joined: 30 May 2008
Posts: 35
Own Kudos [?]: 840 [0]
Given Kudos: 26
walker wrote:
PROBABILITY

This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

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

This topic is included in GMAT ToolKit App (iPhone/iPod Touch)

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

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]
-------------------------------------------------------

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......
Math Expert
Joined: 02 Sep 2009
Posts: 94580
Own Kudos [?]: 643201 [1]
Given Kudos: 86728
1
Bookmarks
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?
Intern
Joined: 21 Mar 2013
Posts: 31
Own Kudos [?]: 560 [0]
Given Kudos: 56
GMAT Date: 03-20-2014
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?
Math Expert
Joined: 02 Sep 2009
Posts: 94580
Own Kudos [?]: 643201 [7]
Given Kudos: 86728
7
Bookmarks
Intern
Joined: 18 Sep 2013
Posts: 7
Own Kudos [?]: 1 [0]
Given Kudos: 18
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?

Math Expert
Joined: 02 Sep 2009
Posts: 94580
Own Kudos [?]: 643201 [2]
Given Kudos: 86728
2
Bookmarks
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?

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
Intern
Joined: 04 Feb 2017
Posts: 9
Own Kudos [?]: 99 [0]
Given Kudos: 197
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?
Manager
Joined: 17 May 2015
Posts: 198
Own Kudos [?]: 3143 [1]
Given Kudos: 85
1
Kudos
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?

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.