GP 2;Probability problem(Four rows of Pegs)

[sharok50]

[img][img]http://www.postimage.org/aV29s7dr.jpg[/img][/img]

Please,help me out.
Thanks in advance

[RonPurewal]

this is a problem in which there aren't that many different possibilities. i'll give a theoretical approach in a sec, just in case there are any readers with that sort of bent, but here's a takeaway for you:
if you see a probability problem in which it's clear that there are relatively few possibilities, then, unless you IMMEDIATELY see a theoretical approach, you should open the problem by LISTING POSSIBILITIES.

whenever you make such a list, the list must of course be organized; if you make a haphazard list of possibilities as they randomly come into your head, with no organizing principle, then you will probably not assemble a complete list (and/or you will accidentally count possibilities more than one time each).

in this case, the ball goes either left or right whenever it hits a peg. this will happen three times between the release and the end of the trajectory, so you have a combination of three L's/R's.

here's an organized list, arranged by increasing numbers of L's (starting with no L's at all, working up to three L's):
LLL
LLR LRL RLL
LRR RLR RRL
RRR

here's another list, arranged alphabetically:
LLL
LLR
LRL
LRR
RLL
RLR
RRL
RRR

either way, you come to the conclusion that there are 8 possibilities, meaning that your probability fraction starts out with a denominator of 8. this is enough to eliminate choice (a), whose denominator is larger than 8, but that's about it.

now, the numerator.
this is the number of possibilities that actually get you into the desired slot.
an examination of the picture reveals that, to get into the desired slot #2, you have to go left exactly 2 times, and right exactly once, in some order. there are three possibilities in the above list that do this: LLR, LRL, and RLL.
therefore, probability = 3 out of 8.

--

THEORY APPROACH:
if you didn't think of this in about 15 seconds, then don't continue to look for it; it's here for your edification only. YOU SHOULD NOT KEEP LOOKING FOR THESE SORTS OF THINGS IF YOU DON'T FIND THEM RIGHT AWAY. REMEMBER THAT YOU SHOULD START LISTING RIGHT AWAY IF YOU DO NOT _IMMEDIATELY_ SEE THE THEORY APPROACH, AS THERE ARE NO POINTS FOR ELEGANCE OR STYLE.

the total number of possibilities is the total number of ways to have three L's or R's.
that's three decisions, with two options (L and R) at each decision.
therefore, the total number of possibilities is 2 x 2 x 2, or 8.

the total number of successes is the total number of ways to have two L's out of three.
this is a combination problem; the number of ways of picking two out of three is 3! / (2!1!), or 3.
alternatively, it's the total number of ways to pick one R out of three, which is clearly 3.

either way, 3 out of 8.

[sharok50]

Thanks Ron for such a great explanation.you are the ultimate GMAT Guru.

[StaceyKoprince]

You're welcome (and I agree - great explanation!)

[sprparvathy]

Hi, Can we also work this out like the following:

An examination of the figure reveals that there are 3 ways to reach cell 2 (by a combination of lefts and rights)

And there are 3 pegs to pass thru in each of these paths. The probability of the ball going either way through this 3 pegs is 1/2 (for each peg).

Therefore, desired probability = 3 * 1/2 *1/2 *1/2 = 3/8.

[mschwrtz]

Yes, sprparvathy, that's sound as well. Notice that Ron divides the number of successful outcomes by the number of total outcomes (probability only comes in after counting), while you take the product of the probabilities. Either works.