, 2020

On each of her four tosses of the coin, Jane will either move up one dot or she will move right one dot.

Since Jane has two possible moves on each of her four tosses of the coin, she has a total of $2\times2\times2\times2=16$ different paths that she may take to arrive at one of $P$, $Q$, $R$, $S$, or $T$. If we denote a move up one dot by $U$, and a move right one dot by $R$, these 16 paths are: $UUUU$, $UUUR$, $UURU$, $UURR$, $URUU$, $URUR$, $URRU$, $URRR$, $RUUU$, $RUUR$, $RURU$, $RURR$, $RRUU$, $RRUR$, $RRRU$, $RRRR$. The probability of tossing a head (and thus moving up one dot) is equal to the probability of tossing a tail (and thus moving right one dot). That is, it is equally probable that Jane will take any one of these 16 paths. Therefore, the probability that Jane will finish at dot $R$ is equal to the number of paths that end at dot $R$ divided by the total number of paths, 16. How many of the 16 paths end at dot $R$? Beginning at $A$, a path ends at $R$ if it has two moves up (two $U$’s), and two moves right (two $R$’s). There are 6 such paths: $UURR$, $URUR$, $URRU$, $RUUR$, $RURU$, $RRUU$. (We note that each of the other 10 paths will end at one of the other 4 dots, $P$, $Q$, $S$, $T$.) After four tosses of the coin, the probability that Jane will be at dot $R$ is $\frac{6}{16}=\frac{3}{8}$.