GMAT Math

GMAT Experts,

Can you plz confirm the best method to approach question as attached below. Please share steps to arrive at correct answer.

anks17 wrote:GMAT Experts,

Can you plz confirm the best method to approach question as attached below. Please share steps to arrive at correct answer.

Both Barker and Carey would be really proud of this question.
Come on down!

Jokes apart, the way I did it is as follows. See the attached figure.

Now, the A, B1, B2, C1,..etc represent the various gaps between the pegs. Since the probability of the ball going left or right = 1/2, we draw and use the decision tree to guide our answer. For each leg of the tree, probability = 1/2. Now the problem is simple. We simply need to count the ways in which the ball reaches number "2" (follow the red arrows!).

For the ball to reach 2, it can do so in only three ways.

1. A->B1->C1-2 : Probability of this event = 1/2 * 1/2 * 1/2 = 1/8
2. A->B1->C2-2 : Probability of this event = 1/8 as well (same as above)
3. A->B2->C2-2 : Probability of this event = 1/8 also.

Total probability of ball reaching 2 = 1/8 + 1/8 + 1/8 = 3/8.


Alternatively, you can just count the paths leading to the four numbers (1,2,3,4) and find the probability from there:

Paths leading from A to 1 = 1 (only A->B1->C1->1 works)
Paths leading from A to 2 = 3 (as we saw above)
Paths leading from A to 3 = 3 (3 is pretty much the symmetrical mirror image of 2)
Paths leading from A to 4 = 1 (4 is similar to 1, A->B2->C3->4 is the only path)

Total number of paths = 1+3+3+1 = 8
Paths leading to 2 = 3
Hence, probability of ball going to 2 = (Paths leading to 2)/(Total number of paths) = 3/8.

Contestant D wins

use the formula-
(4c2)/(2^4)=3/8

can you explain the formula?

avada wrote:can you explain the formula?

Stuart explained it very well -https://www.beatthegmat.com/coin-flip-qu ... 17911.html