19. Jerry starts at the origin $O$ on the number line, and tosses a fair coin 8 times. Each time the coin lands heads, he moves one unit in the positive direction on the number line; when it lands tails, he moves one unit in the negative direction. If the probability that he can reach the point with coordinate 4 is $\frac{a}{b}\left(a, b \in \mathbf{Z}_{+},(a, b)=1\right)$, then the value of $a+b$ is ( ) (for example, if the sequence of coin tosses is “HHTHHHHH”, then he successfully reaches the point 4).
(A) 69
(B) 151
(C) 257
(D) 293
(E) 313
19. B.

Let event $A$ be the event that Jerry can reach the point with coordinates 4.
Consider tossing a coin 8 times.
When there are 4 heads and 4 tails, the only way to reach the 4 point is 1 way.

When there are 5 heads and 3 tails, to reach the 4 point, either the first 4 tosses are all heads, which has $\mathrm{C}_{4}^{1}=4$ ways; or the first 4 tosses have one tail and the 5th toss is a head, which has $\mathrm{C}_{4}^{1}=4$ ways. In total, there are 8 ways.

When there are 6 heads and 2 tails, the number of ways to reach the 4 point is $\mathrm{C}_{8}^{2}=28$ ways.

When there are 7 heads and 1 tail, the number of ways to reach the 4 point is $\mathrm{C}_{8}^{1}=8$ ways.

When there are 8 heads and 0 tails, the number of ways to reach the 4 point is 1 way.
Then $P(A)=\frac{1+8+28+8+1}{2^{8}}=\frac{23}{128}$
$$
\Rightarrow a+b=23+128=151 \text {. }
$$