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Tricky PnC and probablity question

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Tricky PnC and probablity question

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?
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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
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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
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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
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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 :)
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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 ;)
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by sixpointer » Wed Oct 27, 2010 10:29 am
@Shovan bhai can you please explain me probablity question ?
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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.
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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 :)
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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)
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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:)
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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
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