EgmatQuantExpert
The number of paths from one place to another place are as shown in the table below. A, B, C, D are four adjacent points in the given order. If Jack, who is at point A, wants to reach his destination, D. What is the probability that he does not pass through C?

A. \(\frac{1}{4}\)
B. \(\frac{8}{27}\)
C. \(\frac{1}{3}\)
D. \(\frac{3}{7}\)
E. \(\frac{1}{2}\)
Hi,
This question is based on the concept of "the fundamental principle of counting".
TheoryIf there are \(m\) ways to do one thing and \(n\) ways to do another thing, then the total number of ways to do both things is \(m \times n\) ways. This can be further generalized to any number of events. Here, we assume that all the choices are independent of one another.
ProblemAttachment:
E-Gmat_Path.png [ 32 KiB | Viewed 6689 times ]
Following paths are available from A to D:
A-D = 1 path
A-B-D = 2*4 = 8 paths
A-C-D = 3*2 = 6 paths
A-B-C-D = 2*3*2 = 12 paths.
Total no. of paths = 1+8+6+12 = 27 paths. No. of paths without visiting point C = 1+8 = 9 paths.
Required probability = \(\frac{9}{27} = \frac{1}{3}\).
Answer (C).
Thanks.