[THOUGHT] We need to interpret the problem carefully.

We have a natural number a ≤ 6. The experiment: roll a fair die (1-6). If outcome ≤ a, toss coin 5 times, record number of heads. If outcome > a, toss coin 3 times, record number of heads. Repeat 19200 times. Let X be number of times recorded value is 3. Given E(X) = 4800. We need to find P(X ≤ 4800 + 30a). Then compute 1000 * k where k = that probability.

First, find a. Compute E(X) per trial: probability that recorded number of heads = 3.

Let p = probability that recorded heads = 3 in a single experiment.

Case 1: die outcome ≤ a. Probability of this = a/6. Then we toss coin 5 times. Number of heads ~ Binomial(5, 1/2). Probability of exactly 3 heads = C(5,3)*(1/2)^5 = 10/32 = 5/16.

Case 2: die outcome > a. Probability = (6-a)/6. Then toss coin 3 times. Number of heads ~ Binomial(3, 1/2). Probability of exactly 3 heads = C(3,3)*(1/2)^3 = 1/8 = 2/16.

Thus p = (a/6)(5/16) + ((6-a)/6)(2/16) = (1/6)( (5a/16) + (2(6-a)/16) ) = (1/6)( (5a + 12 - 2a)/16 ) = (1/6)*( (3a + 12)/16 ) = (3a + 12)/(96) = (a + 4)/32.

Check: a=0? p = 4/32 = 1/8. That matches case 2 only (since a=0, always >0). For a=6, p = (6+4)/32 = 10/32 = 5/16, matches case 1 only.

Now E(X) = n * p = 19200 * (a+4)/32 = 19200/32 * (a+4) = 600 * (a+4). Given E(X) = 4800, so 600*(a+4) = 4800 => a+4 = 8 => a = 4.

Thus a = 4.

Now we need P(X ≤ 4800 + 30a) = P(X ≤ 4800 + 120) = P(X ≤ 4920).

X is sum of 19200 independent Bernoulli trials, but not identical? Actually each trial is independent, but probability of success is same each trial? Yes, because each trial we roll die and toss coin, but the probability of success (recorded heads = 3) is constant p = (a+4)/32 = (4+4)/32 = 8/32 = 1/4.

Thus X ~ Binomial(n=19200, p=1/4). Expectation μ = np = 4800, variance σ^2 = np(1-p) = 19200*(1/4)(3/4) = 19200(3/16) = 192003/16 = 12003 = 3600? Let's compute: 19200/16 = 1200, times 3 = 3600. So σ = sqrt(3600) = 60.

We need P(X ≤ 4920). Since n is large, approximate by normal distribution with continuity correction? The problem gives standard normal table, so they expect normal approximation without continuity correction? Usually for large n, we can use normal approximation. They ask P(X ≤ 4800 + 30a) = P(X ≤ 4920). Compute z-score: z = (4920 - μ)/σ = (4920 - 4800)/60 = 120/60 = 2.0.

But careful: we need P(X ≤ 4920) = P(Z ≤ 2.0) where Z ~ N(0,1). However, table given is P(0 ≤ Z ≤ z). So P(Z ≤ 2.0) = 0.5 + P(0 ≤ Z ≤ 2.0) = 0.5 + 0.477 = 0.977.

Thus k = 0.977. Then 1000 * k = 977.

But need to verify if continuity correction needed? The problem likely expects normal approximation without continuity correction because they give integer values and table. Also they ask to compute 1000*k, which is integer. So answer likely 977.

Let's double-check calculations with Python to ensure no arithmetic mistakes.

[/THOUGHT] [PYTHON] import math

Given values

n = 19200 E_X = 4800

Solve for a

p = (a+4)/32

E_X = n * p = 19200 * (a+4)/32 = 600*(a+4)

600*(a+4) = 4800 => a+4 = 8 => a = 4

a = 4 print("a =", a)

Compute p

p = (a + 4) / 32 print("p =", p)

Check E_X

E_check = n * p print("E_check =", E_check)

Variance and sigma

var = n * p * (1 - p) sigma = math.sqrt(var) print("sigma =", sigma)

Compute z-score for X <= 4800 + 30a

upper = 4800 + 30 * a print("upper =", upper) z = (upper - E_X) / sigma print("z =", z)

Using given table: P(0 <= Z <= 2.0) = 0.477

P_0_to_z = 0.477 P_Z_le_z = 0.5 + P_0_to_z print("P(X ≤ upper) ≈", P_Z_le_z)

k = P_Z_le_z result = 1000 * k print("1000 * k =", result) [/PYTHON]

[THOUGHT] The Python output confirms that a = 4, p = 0.25, σ = 60, and the z-score is exactly 2.0. Using the given normal table, P(0 ≤ Z ≤ 2.0) = 0.477, so P(Z ≤ 2.0) = 0.5 + 0.477 = 0.977. Thus k = 0.977, and 1000k = 977.0. The problem asks for an integer, so the answer is 977. [/THOUGHT] \boxed{977}