Understanding the Context

In probability theory, ( P(k) ) often represents the probability of achieving exactly ( k ) successes in a fixed number of independent trials, where each trial has two possible outcomes: success or failure. The formula used here follows the Binomial Distribution, a cornerstone of discrete probability.

Breaking Down the Formula:
[
P(2) = inom{4}{2} \left(rac{1}{6}

ight)^2 \left(rac{5}{6} ight)^2
]

• ( inom{4}{2} ) – the binomial coefficient: the number of ways to choose 2 successes out of 4 trials.
  This reflects all possible sequences where exactly 2 successes occur (e.g., SSFF, SFSF, SFFS, FSSF, FSFS, FFS S).

• ( \left(rac{1}{6} ight)^2 ) – probability of 2 successes occurring, each with probability ( rac{1}{6} ).
  This accounts for the specific outcome probability of the successes.

Key Insights

• ( \left(rac{5}{6} ight)^2 ) – probability of 2 failures, each with probability ( rac{5}{6} ).
  Failure probability multiplies across the remaining trials.

Why Use the Binomial Distribution?

The binomial model applies when:

• Trials are independent
  - Each trial has only two outcomes: success or failure
  - The probability of success, ( p = rac{1}{6} ), remains constant
  - Number of trials, ( n = 4 ), is fixed

In this case, we're interested in exactly two successes—whether in quality control, medical trials, survey sampling, or reliability testing—making this a canonical use of the binomial probability formula.

Final Thoughts

Computing the Value Step-by-Step

Let’s calculate the numerical value of ( P(2) ):

1. Compute the binomial coefficient:
   [
   inom{4}{2} = rac{4!}{2!2!} = rac{24}{4} = 6
   ]

2. Compute success part:
   [
   \left(rac{1}{6} ight)^2 = rac{1}{36}
   ]

3. Compute failure part:
   [
   \left(rac{5}{6} ight)^2 = rac{25}{36}
   ]

4. Multiply all components:
   [
   P(2) = 6 \cdot rac{1}{36} \cdot rac{25}{36} = 6 \cdot rac{25}{1296} = rac{150}{1296} = rac{25}{216}
   ]