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5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252
Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25
Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252
Required probability = 1-25/252 = 227/252 _________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
- Bernard Edmonds
Last edited by giddi77 on 07 Jun 2006, 16:20, edited 3 times in total.
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..
it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..
it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.
Prof, question asks what the prob person will NOT travel through C _________________
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..
it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.
Prof, question asks what the prob person will NOT travel through C
oh yes, its E.
the probability cannot be significantly less than 1/2. so A(1/6), B (227/225 = absurd), and C(1/5) are clearly out. the choice is between D and E. if D is not OA, then only E remains.
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..
it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.
Prof, question asks what the prob person will NOT travel through C
oh yes, its E.
the probability cannot be significantly less than 1/2. so A(1/6), B (227/225 = absurd), and C(1/5) are clearly out. the choice is between D and E. if D is not OA, then only E remains.
are the answer choices changing every time. I see that I picked C above, but I wanted to pick B.
Prof - B is 227/252 ... not sure where you got 225 from...maybe it changed from the time you looked. _________________
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..
it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.
Prof, question asks what the prob person will NOT travel through C
oh yes, its E.
the probability cannot be significantly less than 1/2. so A(1/6), B (227/225 = absurd), and C(1/5) are clearly out. the choice is between D and E. if D is not OA, then only E remains.
are the answer choices changing every time. I see that I picked C above, but I wanted to pick B.
Prof - B is 227/252 ... not sure where you got 225 from...maybe it changed from the time you looked.
i am not cuz i do not drink but my condition is more than i am
5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252
Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25
Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252
Required probability = 1-25/252 = 227/252
Giddi - Please elaborate on the logic behind the lines in bold. Thanks!
If you take a simple 2 by 2 path (see picture), From X to Y the shortest path will always have only 4 LEGS (by a LEG I mean either a horizontal or vertical bar) on the picture.
If you denote R = right , D = down, then one of the shortest paths form X to Y could be
R-R-D-D
Similarly, D-D-R-R, D-R-R-D, D-R-D-R, R-D-R-D, R-D-D-R are also shortest possible unique routes. In all there are 6 routes.
Rather than counting all of them we can say that, all possible shortest routes are unique arrangements of 4 letters R, R, D, D
Hence number of routes = 4!/(2!*2!) = 6
In this problem we have 5 Right and 5 Down LEGS.. Hence total routes = 10!/(5!*5!)
Similarly from A->C we have 1-D, and 4-Rs.. Hence total routes from A-> C = 5!/4!
From C-> we again have 1-R and 4-Ds. Hence 5!/4! routes.
From A->B->C = (5!/4!)*(5!/4!)
HTH! BTW when I first got similar problem in my PP test, I couldn't solve it either. I tried to solve this problem for next 2 days and then gave up Found the solution in OG.
Attachments
Shortest_route.GIF [ 1.26 KiB | Viewed 818 times ]
_________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252
Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25
Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252
Required probability = 1-25/252 = 227/252
Giddi - Please elaborate on the logic behind the lines in bold. Thanks!
If you take a simple 2 by 2 path (see picture), From X to Y the shortest path will always have only 4 LEGS (by a LEG I mean either a horizontal or vertical bar) on the picture.
If you denote R = right , D = down, then one of the shortest paths form X to Y could be R-R-D-D Similarly, D-D-R-R, D-R-R-D, D-R-D-R, R-D-R-D, R-D-D-R are also shortest possible unique routes. In all there are 6 routes.
Rather than counting all of them we can say that, all possible shortest routes are unique arrangements of 4 letters R, R, D, D Hence number of routes = 4!/(2!*2!) = 6
In this problem we have 5 Right and 5 Down LEGS.. Hence total routes = 10!/(5!*5!)
Similarly from A->C we have 1-D, and 4-Rs.. Hence total routes from A-> C = 5!/4! From C-> we again have 1-R and 4-Ds. Hence 5!/4! routes.
From A->B->C = (5!/4!)*(5!/4!)
HTH! BTW when I first got similar problem in my PP test, I couldn't solve it either. I tried to solve this problem for next 2 days and then gave up Found the solution in OG.
Thanks a ton giddi! absolutely crystal now . Also reminded me the formula for permutation of n objects wherein a, b.. objects are identicle.
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..
it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.
this is how i was trying to solve this problem (however this is not an elegant approach) but i was drunk (messed up) without drink.
number of ways somebody can pass from A to B through C = 25
total number of ways somebody can pass from A to B = x
number of ways somebody can pass from A to B not through C = x - 25
The prob that somebody can pass from A to B not through C = (x - 25)/x
if we consider on choice B [(252-25)/252 = 227/252), it is exactly what (x - 25)/x is.
Great solution giddi. And the absolute correct one. I cheated myself though. I approximated that from A to B it takes 10 steps and each step there are two options so total number of route is 2^10. Same thing from A to C total number of route is 2^5. So the possibility of passing C is 2^5/2^10 which is 1/2^5, and the possibility of not passing C is 1-1/2^5. I know that the 2^5 and 2^10 are not exact since you don't get two options when you reach the bottom. But I get the sense that the result should be something close to 31/32, or close to 1. And therefore I chose B.
Thought I'd share it too since sometimes this kind of "cheat" or approximation might be useful, especially if you are short on time, or at a loss of what is the correct way to solve a question. _________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.