#### 4Question: A physicist analyzing particle collisions observes that in a series of 12 independent trials, each trial results in one of three outcomes: success (S), partial success (P), or failure (F), with probabilities 0.4, 0.3, and 0.3 respectively. What is the probability that exactly 5 successes, 4 partial successes, and 3 failures occur across the 12 trials?

["### Understanding the Problem: A Multinomial Probability Challenge", "In experimental physics, particularly in high-energy particle collisions, researchers analyze vast numbers of discrete trial events, each yielding one of several possible outcomes. A physicist conducting 12 independent trials observes three distinguishable outcomes in each trial—success (S), partial success (P), and failure (F)—with empirically determined probabilities:\n- ( P(S) = 0.4 )\n- ( P(P) = 0.3 )\n- ( P(F) = 0.3 )", "The physicist seeks the probability of observing exactly 5 successes, 4 partial successes, and 3 failures over the 12 trials. This is a classic problem in probability theory involving the multinomial distribution, which generalizes the binomial distribution to multiple outcomes.", "### The Multinomial Probability Formula", "The probability mass function for a multinomial distribution with ( n ) independent trials and ( k ) possible outcomes is:\n[\nP(X_1 = x_1, X_2 = x_2, \dots, X_k = x_k) = \frac{n!}{x_1! , x_2! , \cdots , x_k!} \cdot p_1^{x_1} p_2^{x_2} \cdots p_k^{x_k}\n]\nwhere:\n- ( n = x_1 + x_2 + \cdots + x_k ) (total number of trials),\n- ( x_i ) is the count of outcome ( i ),\n- ( p_i ) is the probability of outcome ( i ),\n- All ( x_i \geq 0 ) and ( \sum x_i = n ).", "For this problem:\n- ( n = 12 )\n- ( x_S = 5 ), ( x_P = 4 ), ( x_F = 3 )\n- ( p_S = 0.4 ), ( p_P = 0.3 ), ( p_F = 0.3 )", "### Applying the Formula", "Substitute the values into the multinomial formula:", "[\nP(5S, 4P, 3F) = \frac{12!}{5! \cdot 4! \cdot 3!} \cdot (0.4)^5 \cdot (0.3)^4 \cdot (0.3)^3\n]", "Note that ( (0.3)^4 \cdot (0.3)^3 = (0.3)^7 ), so:", "[\nP = \frac{12!}{5! , 4! , 3!} \cdot (0.4)^5 \cdot (0.3)^7\n]", "### Step-by-Step Computation", "1. Compute the multinomial coefficient:\n[\n\frac{12!}{5! , 4! , 3!} = \frac{479001600}{120 \cdot 24 \cdot 6} = \frac{479001600}{17280} = 27720\n]", "2. Compute powers of probabilities:\n[\n(0.4)^5 = 0.01024\n]\n[\n(0.3)^7 = 0.0002187\n]", "3. Multiply all components:\n[\nP = 27720 \cdot 0.01024 \cdot 0.0002187\n]\nFirst, ( 27720 \cdot 0.01024 = 284.0 ) (approx)\nThen, ( 284.0 \cdot 0.0002187 \approx 0.0621 )", "For higher precision:", "[\n27720 \ imes 0.01024 = 284.0028\n]\n[\n284.0028 \ imes 0.0002187 = 0.062048\n]", "### Final Answer", "Thus, the probability that exactly 5 successes, 4 partial successes, and 3 failures occur in 12 independent trials is approximately:\n[\n\boxed{0.06205}\n]\nor about 6.205%.", "This result enables physicists to assess the likelihood of observing such a distribution of outcomes, supporting statistical analysis in particle collision experiments."]