To evaluate the given binomial probability expression:

$^nC_x \cdot p^x \cdot (1 - p)^{n - x}$

where $n = 6$, $p = 0.3$, and $x = 2$, we can break it down step by step.

Step 1: Calculate $^nC_x$ (the binomial coefficient)

The binomial coefficient is given by:

$^nC_x = \frac{n!}{x!(n - x)!}$

Substituting $n = 6$ and $x = 2$:

$^6C_2 = \frac{6!}{2!(6 - 2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$

Step 2: Calculate $p^x$

Now we compute $p^x$:

$p^x = (0.3)^2 = 0.09$

Step 3: Calculate $(1 - p)^{n - x}$

Now, we calculate $(1 - p)^{n - x}$:

$(1 - p)^{n - x} = (1 - 0.3)^{6 - 2} = (0.7)^4 = 0.2401$

Step 4: Multiply all terms

Now we multiply all the terms together:

$15 \cdot 0.09 \cdot 0.2401 = 0.324135$

Final Answer:

The value of the expression is approximately:

$0.324$

Would you like more details or further clarification?

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Relative Questions:

1. How do you derive the binomial coefficient in general for any $n$ and $x$?
2. What is the significance of the $p^x$ term in a binomial probability formula?
3. Why is $(1 - p)^{n - x}$ important in binomial distributions?
4. How would this change if $p = 0.5$ instead of $p = 0.3$?
5. What are some practical examples of binomial distributions?

Tip: The binomial coefficient $^nC_x$ represents the number of ways to choose $x$ successes from $n$ trials.