GMATPrep - PS - Probability

Question

OA later...

freetheking · Answer

Should be D
Total possible ways = 2^3
3 possible ways

Thus, 3/8

gmatornot · Answer

The OS is D... great explaination freetheking !

v1rok · Answer

Total possible ways = 2^3 

Can somebody spoon-feed me why this is true?

freetheking · Answer

step1:　　↙↘
step2:　↙↘↙↘
step3:↙↘↙↘↙↘
　　　A　B 　C 　D
step1: 2^1
step2: 2^2
step3: 2^3
 
if there's nth step, then 2^n

v1rok · Answer

Sorry again. Somehow I just can not seem to grasp this type of problems. I have some kind of mental block or something. I need to get over it...

I can clearly see how we can get only 3 possible ways to get to Cell#2. But my brain refuses to understand why the total number of paths is 8.

From that diagram's above (the one with arrows) I counted 12 arrows. Why then we only have 8 different paths to the bottom.

Is there another way to explain this kind of problem?

v1rok · Answer

Sorry about bumping this. Anyone cares to share an explanation about how to arrive at 2^3 total number of possible ways? It could be something obvious, but I just don't see it...

haas_mba07 · Answer

The way you can look at it is by generating all cases for the three decision points. 

Imagine you have a road which has two choices at every stop. Based on a decision made you will end up in one of the 8 cities. 

Look at the attached figure and should be clearer.

If you think of the pegs as the stops to make a decision, you have a maximum of 8 destinations. (City 1-8).

If you look at the second figure, called "PegDecision", the decision tree is a specialized form with Stops replaced with Pegs. Note that there are fewer pegs in this case. If you count the number of ways of getting to number 2, you will end up with 3.  {Peg A, Peg B, Peg D, 2} {Peg A, Peg B, Peg F, 2} {Peg A, Peg C, Peg F, 2}

v1rok · Answer

Thank you, haas! It made it a little clearer, but not completely. On a subconcsious level I kind of understand how some cities can merge together: you drive to Dallas, TX, but you also arrive at Fort Worth at the same time. O-oh, actually this analogy just hit me! Yeah, it is exactly like this: just choose two different roads that lead to the same place. I think I got it! Thanks!!!

haas_mba07 · Answer

v1rok, 
 I believe the correct way to solve this problem is as below. I was thinking about this and the 3 decision points are independent events. The overall probabibility of getting the ball in slot 2 is :

As mentioned before there are three ways of getting the ball in slot 2 : 
{Peg1 Left, Peg2 Left, Peg3 Right}.......(1)
{Peg1 Left, Peg2 Right, Peg3 Left}.......(2)
{Peg1 Left, Peg2 Right, Peg3 Right}......(3)

All the individual peg decisions are independent events.


P(Ball in Slot 2) = Prob(Event (1)) + Prob(Event (2)) + Prob(Event (3))
= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2
= 1/8 + 1/8 + 1/8 
= 3/8

The fact that 2^3 =8 has nothing to do with it!!

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