Probability with rows of pegs

[megamix06]

Hello, this is my first post. As I found this forum very helpful I can find answer for my hardest question problems.

[Stuart@KaplanGMAT]

Hi!

Although it may not look like it at first glance, this is really a coin flip question! It may seem a lot more complicated, but it's actually the exact same question as:

If you flip a fair coin three times, what's the probability of getting exactly 2 heads?

The first thing to note is that we have a 50/50 chance each time we hit a peg of going left or right. When you see 50/50, think coin flip!

First, our general equation:

Probability = # desired outcomes / total # of possibilities.

We're hitting 3 pegs on our way to our final resting spot. Since each peg leads to two possibilities (left and right), there are going to be 2^3 = 8 possible ways we can get to the bottom from the top.

To get to "2", we have to go left twice and right once. There are 3 ways this could happen: LLR, LRL and RLL.

So, the # of desired outomes is 3 and the total number of possibilities is 8. Accordingly, there's a 3/8 chance of landing in the "2": choose (d).

* * *

To fill out the rest of the puzzle:

Landing in "3" is analogous to landing in "2"; we have to go right twice and left once, again giving us 3 desired patterns: LRR, RLR, RRL.

To get to the "1" or "4" spots, we have to go left or right every time.

To get to "1", we have to go L L L on each peg. Since there's only one way to go LLL, there's a 1/8 chance of landing in "1". Landing in "1" is the equivalent of tossing 3 heads!

To get to "4", we have to go RRR on each peg. Since there's only one way to go RRR, there's a 1/8 chance of landing in "4". Landing in "4" is the equivalent of tossing 3 tails!

[megamix06]

Thanks Stewart. I didn't think that this could be solved using coin flipping. Seemed much harder !