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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Tricky PnC and probablity question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times
by sixpointer » Wed Oct 27, 2010 5:53 am
1.AMS employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years is 0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9 .The probability that after 10 years at least 6 of them still at work?


2.6 people go to a temple.While returning none of them is wearing their own footwear.
In how many ways is this possible?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 52
Joined: Thu Feb 25, 2010 9:08 am
Thanked: 5 times

by akdayal » Wed Oct 27, 2010 6:35 am
My solution:
1. 1 - P(only one remain after 10 years) - p(no one remains after 10 years)

P(no one remains after 10 years) = 0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.1
P(1 remains after 10 years) = 0.2*0.7*0.6*0.5*0.4*0.3*0.2*0.1 + 0.8*0.3*0.6*0.5*0.4*0.3*0.2*0.1 + 0.8*0.7*0.4*0.5*0.4*0.3*0.2*0.1 + 0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.1 + 0.8*0.7*0.6*0.5*0.6*0.3*0.2*0.1 + 0.8*0.7*0.6*0.5*0.4*0.7*0.2*0.1 + 0.8*0.7*0.6*0.5*0.4*0.3*0.8*0.1 + 0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.9

But certainly, calculation will take a lot of time. I'm not sure this is perfect method.

2. 6![1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6!] = 505
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Wed Oct 27, 2010 7:01 am
akdayal wrote:My solution:
1. 1 - P(only one remain after 10 years) - p(no one remains after 10 years)

P(no one remains after 10 years) = 0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.1
P(1 remains after 10 years) = 0.2*0.7*0.6*0.5*0.4*0.3*0.2*0.1 + 0.8*0.3*0.6*0.5*0.4*0.3*0.2*0.1 + 0.8*0.7*0.4*0.5*0.4*0.3*0.2*0.1 + 0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.1 + 0.8*0.7*0.6*0.5*0.6*0.3*0.2*0.1 + 0.8*0.7*0.6*0.5*0.4*0.7*0.2*0.1 + 0.8*0.7*0.6*0.5*0.4*0.3*0.8*0.1 + 0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.9

But certainly, calculation will take a lot of time. I'm not sure this is perfect method.

2. 6![1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6!] = 505

for 2nd one 265 is the answer :) could you please tell me the logic
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Wed Oct 27, 2010 7:26 am

by reddy43 » Wed Oct 27, 2010 7:29 am
for n things not to be in their right places
no. of ways = n!{1/2!-1/3!+....(-1)^n*1/n!} where n>=2
so we will have this one as 265 ways
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Wed Oct 27, 2010 7:41 am
reddy43 wrote:for n things not to be in their right places
no. of ways = n!{1/2!-1/3!+....(-1)^n*1/n!} where n>=2
so we will have this one as 265 ways

can u explain me how does this formula work ? what is the logic behind this? where to use :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Wed Oct 27, 2010 8:16 am
sixpointer wrote:
reddy43 wrote:for n things not to be in their right places
no. of ways = n!{1/2!-1/3!+....(-1)^n*1/n!} where n>=2
so we will have this one as 265 ways

can u explain me how does this formula work ? what is the logic behind this? where to use :)
This is a Q of Derangement :)

I can explain how it works but do not ever dare to get into the logic as the concept behind the formula will drive you crazy...

Derangement means the item should not appear at the place where it originally was.

Say ABC now we can arrange this in 3!=6 ways
1.ABC
2.ACB
3.BAC
4.BCA
5.CAB
6.CBA

But the derangement will be 2 only BCA and CAB. see none of A B or C are being placed at their original places. A cant be at 1st, B cant be at 2nd and C cant be at 3rd place.

So just remember the formula no. of ways = n!{1/2!-1/3!+....(-1)^n*1/n!} where n>=2 .

My example n=3 Thus, 3!{1/2! - 1/3!} = 2.

Need more info https://en.wikipedia.org/wiki/Derangement
but it will have Limit (Math). So be careful ;)
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Wed Oct 27, 2010 10:29 am
@Shovan bhai can you please explain me probablity question ?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Wed Oct 27, 2010 10:43 am
sixpointer wrote:@Shovan bhai can you please explain me probablity question ?
If your answer is incorrect then I believe you have to subtract 6 instead of 3. (347,357,367, And 763,753,743) calculated twice. Once from 3 to 10 and other from 7 to 3.
Question is asking minimum is 3 or maximum is 7
So in these 6 cases the case is minimum is 3 AND maximum is 7.

Rest all approach is correct I believe.
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Wed Oct 27, 2010 10:51 am
shovan85 wrote:
sixpointer wrote:@Shovan bhai can you please explain me probablity question ?
If your answer is incorrect then I believe you have to subtract 6 instead of 3. (347,357,367, And 763,753,743) calculated twice. Once from 3 to 10 and other from 7 to 3.
Question is asking minimum is 3 or maximum is 7
So in these 6 cases the case is minimum is 3 AND maximum is 7.

Rest all approach is correct I believe.

@Shovan try it :- Three numbers are chosen at random without replacement from( 1,2,3...10).The probability that the minimum is 3 or the maximum is seven?


P.S :- I got my mistake :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Wed Oct 27, 2010 10:56 am
sixpointer wrote:
shovan85 wrote:
sixpointer wrote:@Shovan bhai can you please explain me probablity question ?
If your answer is incorrect then I believe you have to subtract 6 instead of 3. (347,357,367, And 763,753,743) calculated twice. Once from 3 to 10 and other from 7 to 3.
Question is asking minimum is 3 or maximum is 7
So in these 6 cases the case is minimum is 3 AND maximum is 7.

Rest all approach is correct I believe.

@Shovan try it :- Three numbers are chosen at random without replacement from( 1,2,3...10).The probability that the minimum is 3 or the maximum is seven?


P.S :- I got my mistake :)
Is it 1/10? (My answer includes minimum is 3 AND maximum is 7 also)
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Wed Oct 27, 2010 11:03 am
shovan85 wrote:
sixpointer wrote:
shovan85 wrote:
sixpointer wrote:@Shovan bhai can you please explain me probablity question ?
If your answer is incorrect then I believe you have to subtract 6 instead of 3. (347,357,367, And 763,753,743) calculated twice. Once from 3 to 10 and other from 7 to 3.
Question is asking minimum is 3 or maximum is 7
So in these 6 cases the case is minimum is 3 AND maximum is 7.

Rest all approach is correct I believe.

@Shovan try it :- Three numbers are chosen at random without replacement from( 1,2,3...10).The probability that the minimum is 3 or the maximum is seven?


P.S :- I got my mistake :)
Is it 1/10? (My answer includes minimum is 3 AND maximum is 7 also)

it should not include 3 and max 7 case , See the bold Part:)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Mon Oct 18, 2010 9:04 am
Thanked: 2 times

by sixpointer » Wed Oct 27, 2010 12:07 pm
one more :- A, B, C and D are four towns any three of which are non-collinear. Then the number of ways to construct
three roads each joining a pair of towns so that the roads do not form a triangle is

9 , 10 , 12 , 24
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1