[THOUGHT] Let me analyze this problem step by step.

First, I need to find the value of $a$ given that $E(X) = 4800$.

When we roll a die:

• With probability $\frac{a}{6}$, the outcome is $\leq a$, and we toss the coin 5 times
• With probability $\frac{6-a}{6}$, the outcome is $> a$, and we toss the coin 3 times

The probability of getting exactly 3 heads:

• In 5 tosses: $\binom{5}{3} \cdot \left(\frac{1}{2}\right)^5 = 10 \cdot \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$
• In 3 tosses: $\binom{3}{3} \cdot \left(\frac{1}{2}\right)^3 = 1 \cdot \frac{1}{8} = \frac{1}{8}$

So the probability of recording 3 in one experiment is: $p = \frac{a}{6} \cdot \frac{5}{16} + \frac{6-a}{6} \cdot \frac{1}{8}$

$p = \frac{5a}{96} + \frac{6-a}{48} = \frac{5a}{96} + \frac{2(6-a)}{96} = \frac{5a + 12 - 2a}{96} = \frac{3a + 12}{96} = \frac{a + 4}{32}$

Since $X$ follows a binomial distribution with $n = 19200$ trials and probability $p$: $E(X) = n \cdot p = 19200 \cdot \frac{a + 4}{32} = 600(a + 4)$

Given $E(X) = 4800$: $600(a + 4) = 4800$ $a + 4 = 8$ $a = 4$

Now I need to find $P(X \leq 4800 + 30a) = P(X \leq 4800 + 120) = P(X \leq 4920)$

For the variance: $Var(X) = n \cdot p \cdot (1-p) = 19200 \cdot \frac{8}{32} \cdot \frac{24}{32} = 19200 \cdot \frac{1}{4} \cdot \frac{3}{4} = 19200 \cdot \frac{3}{16} = 3600$

So $\sigma = \sqrt{3600} = 60$

Using normal approximation: $Z = \frac{X - \mu}{\sigma} = \frac{X - 4800}{60}$

$P(X \leq 4920) = P\left(Z \leq \frac{4920 - 4800}{60}\right) = P(Z \leq 2)$

$P(Z \leq 2) = 0.5 + P(0 \leq Z \leq 2) = 0.5 + 0.477 = 0.977$

So $k = 0.977$ and $1000 \times k = 977$

Let me verify this with Python. [/THOUGHT] [PYTHON] from math import comb

First, find a given E(X) = 4800

P(3 heads in 5 tosses)

p_5 = comb(5, 3) * (1/2)**5 print(f"P(3 heads in 5 tosses) = {p_5} = {comb(5,3)}/32")

P(3 heads in 3 tosses)

p_3 = comb(3, 3) * (1/2)**3 print(f"P(3 heads in 3 tosses) = {p_3} = 1/8")

For a given a, probability of recording 3 in one experiment

p = (a/6) * p_5 + ((6-a)/6) * p_3

p = (a/6) * (5/16) + ((6-a)/6) * (1/8)

p = 5a/96 + (6-a)/48 = 5a/96 + 2(6-a)/96 = (5a + 12 - 2a)/96 = (3a + 12)/96 = (a+4)/32

E(X) = 19200 * p = 19200 * (a+4)/32 = 600(a+4)

600(a+4) = 4800

a + 4 = 8

a = 4

a = 4 p = (a + 4) / 32 print(f"a = {a}") print(f"p = {p} = {a+4}/32")

Verify E(X)

n = 19200 E_X = n * p print(f"E(X) = {E_X}")

Variance

Var_X = n * p * (1 - p) print(f"Var(X) = {Var_X}") sigma = Var_X ** 0.5 print(f"sigma = {sigma}")

P(X <= 4800 + 30a) = P(X <= 4800 + 120) = P(X <= 4920)

threshold = 4800 + 30 * a print(f"threshold = {threshold}")

Z-score

z = (threshold - E_X) / sigma print(f"z = {z}")

P(Z <= 2) = 0.5 + P(0 <= Z <= 2) = 0.5 + 0.477

k = 0.5 + 0.477 print(f"k = P(X <= 4920) = {k}")

1000 * k

result = 1000 * k print(f"1000 * k = {result}") [/PYTHON]

[THOUGHT] The Python calculation confirms my analysis:

• We found $a = 4$ from the condition $E(X) = 4800$
• The probability $p = \frac{a+4}{32} = \frac{8}{32} = \frac{1}{4}$
• Variance is $3600$, so $\sigma = 60$
• We need $P(X \leq 4920)$, which corresponds to $z = 2$
• Using the standard normal table: $P(Z \leq 2) = 0.5 + 0.477 = 0.977$
• Therefore $k = 0.977$ and $1000 \times k = 977$ [/THOUGHT]

\boxed{977}