Sure! Let's go through this step-by-step.
Step 1: Understanding the Binomial Distribution Formula
The binomial distribution formula is given by:
[tex]\[ P(r) = C(n, r) \cdot p^r \cdot (1 - p)^{n - r} \][/tex]
where:
- [tex]\( P(r) \)[/tex] is the probability of having exactly [tex]\( r \)[/tex] successes in [tex]\( n \)[/tex] trials.
- [tex]\( C(n, r) \)[/tex] is the binomial coefficient, which describes the number of ways to choose [tex]\( r \)[/tex] successes out of [tex]\( n \)[/tex] trials, and is calculated as [tex]\( \frac{n!}{r!(n-r)!} \)[/tex]
- [tex]\( p \)[/tex] is the probability of success on a single trial.
- [tex]\( (1 - p) \)[/tex] is the probability of failure on a single trial.
- [tex]\( n \)[/tex] is the total number of trials.
- [tex]\( r \)[/tex] is the number of successes we are interested in.
Step 2: Calculate the Binomial Coefficient [tex]\( C(n, r) \)[/tex]
Given [tex]\( n = 100 \)[/tex] and [tex]\( r = 2 \)[/tex], the binomial coefficient [tex]\( C(n, r) \)[/tex] is:
[tex]\[ C(100, 2) = \frac{100!}{2!(100-2)!} \][/tex]
This can be computed as:
[tex]\[ C(100, 2) = \frac{100 \times 99}{2 \times 1} = 4950 \][/tex]
Step 3: Compute the Probability [tex]\( P(r) \)[/tex]
Next, we need to plug in the values into the binomial distribution formula. Given [tex]\( p = 0.03 \)[/tex], [tex]\( n = 100 \)[/tex], and [tex]\( r = 2 \)[/tex], we get:
[tex]\[ P(2) = C(100, 2) \cdot (0.03)^2 \cdot (1 - 0.03)^{100 - 2} \][/tex]
[tex]\[ P(2) = 4950 \cdot (0.03)^2 \cdot (0.97)^{98} \][/tex]
Let's break it down:
1. Compute [tex]\( (0.03)^2 = 0.0009 \)[/tex]
2. Compute [tex]\( (0.97)^{98} \approx 0.0452 \)[/tex]
Now, multiply these together along with the binomial coefficient:
[tex]\[ P(2) = 4950 \cdot 0.0009 \cdot 0.0452 \approx 0.2252 \][/tex]
Step 4: Rounding the Final Answer
Finally, we round our result to four decimal places:
[tex]\[ P(2) \approx 0.2252 \][/tex]
So, for [tex]\( n = 100 \)[/tex], [tex]\( p = 0.03 \)[/tex], and [tex]\( r = 2 \)[/tex], the probability [tex]\( P(r) \)[/tex] is approximately 0.2252.
