BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability questions

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 145
Joined: Sun Mar 15, 2015 7:27 am
Thanked: 2 times
Followed by:1 members

Probability questions

by src_saurav » Fri Apr 03, 2015 9:19 pm
Hi,

Please help me with the below

1)Judges will select 5 from 7.They will then rank and give prizes to 3 highest ranked contestants: blue,red and yellow.How many arrangements are possible?

options
10;21;210;420;1260

My answer
7c3=35 but t is not the option.PLease help and explain.



2)integers 1 to 100 are each written on a slip of paper inclusive and one slip is drwan out without replacement.What is the probablity that it is neither even nor multiple of 3.

options
83;67;50;33;17


my answer

got stuck

1-((50/100)+(33/100)-P(even and multiples of 3))

please help
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sat Apr 04, 2015 2:28 am
The first problem should read as follows:
Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible?

10
21
210
420
1,260
The selection of 5 finalists is IRRELEVANT and has no bearing on the number of possible arrangements.

Number of options for blue = 7. (Any of the 7 contestants could be ranked blue.)
Number of options for red = 6. (Any of the 6 remaining contestants could be ranked red.)
Number of options for yellow = 5. (Any of the 5 remaining contestants could be ranked yellow.)
To combine these options, we multiply:
7*6*5 = 210.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sat Apr 04, 2015 2:43 am
The second problem should read as follows:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, what is the probability that the number on it is neither even nor a multiple of 3?

83%
67%
50%
33%
17%
One approach:
Total integers = (evens) + (multiples of 3) - (integers that are BOTH even AND divisible by 3) + (neither).

Total integers = 100.

Since half of the integers between 1 and 100 inclusive are even, the number of EVEN integers = 50.

Since 100/3 = 33.33, the number of multiples of 3 between 1 and 100 inclusive = 33.

An integer that is BOTH even AND divisible by 3 is a MULTIPLE OF 6.
Since 100/6 = 16.66, the number of multiples of 6 between 1 and 100 inclusive = 16.

Plugging these values into the equation above, we get:
100 = 50 + 33 - 16 + N
N = 33.

Thus:
P(neither even nor a multiple of 3) = 33/100 = 33%.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
User avatar
Legendary Member
Posts: 2663
Joined: Wed Jan 14, 2015 8:25 am
Location: Boston, MA
Thanked: 1153 times
Followed by:128 members
GMAT Score:770

by DavidG@VeritasPrep » Sat Apr 04, 2015 2:55 am
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, what is the probability that the number on it is neither even nor a multiple of 3?

83%
67%
50%
33%
17%
An alternative is to list out numbers and try to detect a pattern:

1, 2, 3 ---> 1/3 not even or multiple of 3
4, 5, 6---> 1/3 not even or multiple of 3
7, 8, 9, ---> 1/3 not even or multiple of 3

1/3 = 33%
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Apr 04, 2015 6:22 am
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, what is the probability that the number on it is neither even nor a multiple of 3?

83%
67%
50%
33%
17%
As David says, we can list and look for a pattern.
ALSO NOTE that, even if we don't spot a pattern, the answers are spread out enough to still find the correct answer. Here's what I mean:

Check the first 20 integers and list those that are neither even nor a multiple of 3. We get: 1, 5, 7, 11, 13, 17, 19 (7 values)
So, 7 of the first 20 integers meet the given criteria.
7/20 = 35%, so B looks like the only candidate that's reasonably close.

If you're not convinced, try checking the first 40 integers instead. We get: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37 (13 values)
So, 13 of the first 40 integers meet the given criteria.
13/40 is CLOSE TO 13/39 (which = 1/3), so B is still the only candidate that's reasonably close.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Apr 04, 2015 6:28 am
Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible?

10
21
210
420
1,260
My approach is the same as Mitch's, with a few extra bits filled in.


Take the task of assigning the ribbons and break it into stages.

Stage 1: Select the person to receive the first place (blue) ribbon
Since there are 7 people to choose from, we can complete stage 1 in 7 ways

Stage 2: Select the person to receive the second place (red) ribbon
Since there are now 6 people remaining to choose from, we can complete stage 2 in 6 ways

Stage 3: Select the person to receive the third place (yellow) ribbon
Since there are now 6 people remaining to choose from, we can complete stage 2 in 5 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus assign all 3 ribbons) in (7)(6)(5) ways ([spoiler]= 210 ways[/spoiler])

Answer: C
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 145
Joined: Sun Mar 15, 2015 7:27 am
Thanked: 2 times
Followed by:1 members

by src_saurav » Sat Apr 04, 2015 6:41 am
regarding the judge question.it looks like you have made it a permutation question.I did not understand why order would matter here.As i use FCP when order matters and combo when it doesn't.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat Apr 04, 2015 7:14 am
src_saurav wrote:regarding the judge question.it looks like you have made it a permutation question.I did not understand why order would matter here.As i use FCP when order matters and combo when it doesn't.
Hi src_saurav,

In your original post, you suggested that the answer might be 7C3 (which = 35)
This is a good start.
7C3 represents the number of ways to select 3 people from 7 people (if order does not matter).

NOTE: When I say that order doesn't matter, I'm saying that being selected 1st is no different from being selected 2nd or 3rd)

Your calculations to this point are perfectly fine.
HOWEVER, now that you've selected 3 people (where order does not matter), who gets the blue ribbon? Or the red ribbon? Or the yellow one? So, we still have some work to do.
Assign someone the blue ribbon: We can give it to any of the 3 selected people. So, we can do this in 3 ways.
Assign someone the red ribbon: We can give it to either of the 2 remaining people. So, we can do this in 2 ways.
Assign someone the yellow ribbon: We must give it to the last remaining person. So, we can do this in 1 way.

So, the total number of possibilities = (35)(3)(2)(1)
= 210
= C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Master | Next Rank: 500 Posts
Posts: 111
Joined: Sat Mar 07, 2015 11:00 pm
Thanked: 8 times
Followed by:1 members

by binit » Sat May 02, 2015 9:54 pm
Hi everybody,

The judges Qn, though an easy Qn, was actually very interesting.
What I did right after seeing the problem: N = 7C5*5C3 = 210
Then I thought why should 5 matter at all, I did: N = 7C3*3! = 210
After some moments I tried: N = 7P3 = 210 (since every constant gets a different colored ribbon, order matters, assigning ribbon is just like numbering)
With FCP, as Mich said, N = 7*6*5 = 210 (easiest this way)
I would like to know whether there is any generalization, towards which we are approaching.. Experts' comments needed.

~Binit.
Top
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Sun May 03, 2015 10:36 pm
binit wrote:Hi everybody,

The judges Qn, though an easy Qn, was actually very interesting.
What I did right after seeing the problem: N = 7C5*5C3 = 210
Then I thought why should 5 matter at all, I did: N = 7C3*3! = 210
After some moments I tried: N = 7P3 = 210 (since every constant gets a different colored ribbon, order matters, assigning ribbon is just like numbering)
With FCP, as Mich said, N = 7*6*5 = 210 (easiest this way)
I would like to know whether there is any generalization, towards which we are approaching.. Experts' comments needed.

~Binit.
Your first approach worked because for these particular integers (n choose k) * (k choose r) = n permute (n - r). (Try the problem with n = 8 people instead of n = 7, however.)

Your second and third approaches gave the same answer because for any valid n and k (n choose k) * k! = (n permute k).

Your fourth answer worked because it IS the permutations formula, just written in a format that happens work nicely when n and (n - k) are very close. (That is, 7 permute 3 is perhaps better written as 7*6*5, while 70 permute 30 probably should be left as is.)
Top
Master | Next Rank: 500 Posts
Posts: 111
Joined: Sat Mar 07, 2015 11:00 pm
Thanked: 8 times
Followed by:1 members

by binit » Mon May 04, 2015 11:36 pm
Your first approach worked because for these particular integers (n choose k) * (k choose r) = n permute (n - r). (Try the problem with n = 8 people instead of n = 7, however.)

Your second and third approaches gave the same answer because for any valid n and k (n choose k) * k! = (n permute k).

Your fourth answer worked because it IS the permutations formula, just written in a format that happens work nicely when n and (n - k) are very close. (That is, 7 permute 3 is perhaps better written as 7*6*5, while 70 permute 30 probably should be left as is.)
Thanks Matt,

So, I was just lucky and basically it is a permutation problem and we have to identify that 5 is irrelevant here.
Would you pls comment on how to decide whether FCP or Formula might be the easiest/ fastest approach to a counting problem?

~Binit.
Top
Master | Next Rank: 500 Posts
Posts: 137
Joined: Fri Nov 13, 2015 11:01 am
Thanked: 1 times
Followed by:2 members

by Amrabdelnaby » Sun Nov 15, 2015 6:47 am
Hi Brent,

Could you please tell me whats wrong with my thinking process?

P(not even or not not divisible by 3) = p(not even) = p(not even) + p (not divisble by 3) - P ( not even & not divisible by 3)

p(not even) = 1/2
p(not divisible by 3) = 1/3
p(not even and not divisible by 3) = p(not even) x (p not divisible by 3) = 1/2 x 1/3 = 1/6

therefore: P(not even or not not divisible by 3) = 1/2 + 1/3 - 1/6 = 4/6 = 66.6%
GMATGuruNY wrote:The second problem should read as follows:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, what is the probability that the number on it is neither even nor a multiple of 3?

83%
67%
50%
33%
17%
One approach:
Total integers = (evens) + (multiples of 3) - (integers that are BOTH even AND divisible by 3) + (neither).

Total integers = 100.

Since half of the integers between 1 and 100 inclusive are even, the number of EVEN integers = 50.

Since 100/3 = 33.33, the number of multiples of 3 between 1 and 100 inclusive = 33.

An integer that is BOTH even AND divisible by 3 is a MULTIPLE OF 6.
Since 100/6 = 16.66, the number of multiples of 6 between 1 and 100 inclusive = 16.

Plugging these values into the equation above, we get:
100 = 50 + 33 - 16 + N
N = 33.

Thus:
P(neither even nor a multiple of 3) = 33/100 = 33%.

The correct answer is D.
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sun Nov 15, 2015 7:55 am
Amrabdelnaby wrote:Could you please tell me whats wrong with my thinking process?

P(not even or not not divisible by 3) = p(not even) = p(not even) + p (not divisble by 3) - P ( not even & not divisible by 3)
neither X nor Y means not X AND not Y.
In the question stem, the phrase neither even nor a multiple of 3 means not even AND not a multiple of 3.
In your solution, the or in red misrepresents what the question stem is asking.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Post new topic Post Reply
• Page 1 of 1