IF there are 5 people (A,B,C,D,E) running what is the

Question

IF there are 5 people (A,B,C,D,E) running what is the probability that A always finishes before C.

I figured the answer is 
A can finish before B in 10 ways 
For each of these ways C,D,E can be arranged in 6 ways

total for =60
total = 5! =120
probability = 1/2

but that cant be correct (wuld be if there were only a and b)

is the ans 10/120?

stuti · Answer

here's my take: 59/60

A will NOT finish before C in only 2 cases:
Case 1 - C finishes first
Case 2 - A finishes last 

If there were no restrictions, total number of ways in which all 5 could be arranged from 1->5 = 5! = 120

So P(Afinishing before C) = 1 - P(A not finishing before C)
 = 1 - 2/120
 = 59/60

what do others say? and whats the OA?

hardworker_indian · Answer

Yes, the answer is 1/2.

There are two possibilities: either A finishes before C or after C. So, the probability for each of this event = 1/2

venksune · Answer

These are the possible scenarios wherein A is before C.
1. A C _ _ _. Here, Apart from A and C the remaining three guys can follow in 3 x 2 x 1 ways = 6 ways. 
2. _ A C _ _ . Here The first can be any of the three guys (so 3 ways) - Followed by A and C and then any of the two guys and then the last guy. Again 6 ways.
3. _ _A C _. 6 ways here too.
4. _ _ _ A C. 6 ways here too.

So we have 6 *4 = 24 ways.

Of which AC can be in the above order in 4 ways.

So the probability is 4/24 = 1/6 ways.

Whats' the OA? Am I reading the question wrong? Pls calrify.

ywilfred · Answer

A is first, and C is 2nd, 3rd 4th or 5th-->4 combinations (we don't care about placing of B,D,E)
A is second and C is 3rd, 4th or 5th --> 3 combinations
A is 3rd, C is 4th or 5th --> 2 combinations
A is 4th, C is 5th --> 1 combination

Total: 10 combinations

Total number of combinations is the number of ways to arrange 5 people --> 5! = 120

So P= 10/120 = 1/12

ywilfred · Answer

venksune, do we have to bother about the placing of B,D and E ?

If A comes in 1st, and C second

A,C, _, _, _ --> I think we don't have to bother about B,D and E. What we're intersted is the number of ways C comes in later. A,C,B,D,E and A,C,D,E,B etc all are 1 way, that is, A is first, C is second.

venksune · Answer

My My, I think I had read the question wrongly before. A finishes before C and not just before C. So, this changes the logic a little bit. Thanks to ywilfred pointing that out...let me start over...

We have..
1. A _ _ _ _. A and others in 4! ways = 24 ways.
2. _ A _ _ _. The first guy can be any of B, D and E 3 ways (cant be C). The other three slots can be any of the remaining 3 guys in 3! ways. So total of 18 ways.
3. _ _ A _ _. The first two guys can be 3*2 ways and the remaining 2 slots can be 2*1 ways . Totalling to 6 * 2 ways = 12 ways.
4. _ _ _ AC. The first three slots can be in 3! = 6 ways.

We now have a total of 24+18+12+6 = 60 ways.
Off which A is before C in 4 ways.

So the probability s 4/60 = 1/15. 

ywilfred there is still some discrepancy in our results. Are we missing anything here?

twixt · Answer

This problem is a symetric one
There is no need to dive into deep calculation event if final result should be the same : Prob (A > C) = Prob (C > A)
So Answer = 1/2

The calculation is :
number of combinations : 5!
Various ways A is better ranked than C : 10, remaining combinations for the 3 free places : 3!

=3!*10/5!=1/2

hardworker_indian · Answer

Yes, agree with twixt. Whenever we see a too complicated pb problem, see if the following applies:
1. Symmetry rule: Pb of an event happening and not happening are equal, then the pb = 1/2
2. Negative rule: If the probability of an event not happening is easy to solve, then pb = 1 - (pb of event not happening)

Paul · Answer

I agree with your reasoning until the part in blue. You have 60 ways  in which A is ahead of C . Total number of ways to finish race: 5! = 120. You got the answer right there! 60/120 = 1/2

venksune · Answer

Thanks Paul... what am i doing..

IF there are 5 people (A,B,C,D,E) running what is the

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