Recently, for the sake of variety, Mr. P decided that at each intersection
where he had a choice, he would flip a "fair coin" to determine whether he would
go north (Heads) or east (Tails).
If Mr. P is walking from intersection A to meet Mrs. P at intersection L,
what is the probability that he will pass through intersection H?
Remember that the probability of Heads or Tails on a fair coin is 1/2.
One possible path is shown. It resulted from the four tosses HTHH.
Note that there is no choice on the fifth move,
so no toss is needed.
Solution to the Problem:
The probability that Mr. P will pass through intersection H is 5/16.
Use the diagram below.
If Mr. P starts at A on his way to L, then the probability that he gets to B is 1/2,
since he flips the fair coin at A. Likewise, by symmetry, the probability that he gets to E is also 1/2.
The probability (P) that he gets to F is the sum of the probabilities of coming from B and from E.
For example, P(F from B) = 1/2 * P(B) = 1/2 * 1/2 or 1/4.
Likewise, P(F from E) is also 1/4. Thus, P(F), the probability that he gets to F from B or E, is 1/4 + 1/4 = 1/2.
In a similar fashion, P(C) = P(I) = 1/4. Since there is no choice of direction in going from I to J,
P(J from I) = 1 * P(I) = 1 * 1/4 = 1/4.
Here are the probabilities for this grid. Note that the probability that Mr. P will meet Mrs. P is 1.
| P(D) = 1/8 | P(H) = 5/16 | P(L) = 1 |
| P(C) = 1/4 | P(G) = 3/8 | P(K) = 11/16 |
| P(B) = 1/2 | P(F) = 1/2 | P(J) = 1/2 |
| P(A) = 1 | P(E) = 1/2 | P(I) = 1/4 |
Correctly solved by:
| 1. David & Judy Dixon | Bennettsville, South Carolina |