Sure, let’s find the probability \(P(X = 28)\) in a binomial distribution given the parameters \( n = 30 \), \( p = 0.97 \), and \( x = 28\).
We will use the binomial probability formula:
[tex]\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \][/tex]
where:
- \( \binom{n}{x} \) is the binomial coefficient
- \( p \) is the probability of success on a single trial
- \( 1-p \) is the probability of failure on a single trial
1. Calculate the binomial coefficient \( \binom{n}{x} \):
[tex]\[
\binom{30}{28} = \frac{30!}{28! \cdot (30-28)!} = \frac{30 \times 29}{2 \times 1} = 435
\][/tex]
2. Calculate \( p^x \):
[tex]\[
p = 0.97, \quad x = 28
\][/tex]
[tex]\[
p^x = 0.97^{28} = 0.42619520516862314
\][/tex]
3. Calculate \( (1-p)^{n-x} \):
[tex]\[
1-p = 0.03, \quad n-x = 30-28 = 2
\][/tex]
[tex]\[
(1-p)^{n-x} = 0.03^2 = 0.0009000000000000016
\][/tex]
4. Combine these values to find the binomial probability \( P(X = 28) \):
[tex]\[
P(X = 28) = \binom{30}{28} \cdot 0.97^{28} \cdot 0.03^2
\][/tex]
[tex]\[
P(X = 28) = 435 \times 0.42619520516862314 \times 0.0009000000000000016
\][/tex]
[tex]\[
P(X = 28) = 0.16685542282351626
\][/tex]
Therefore, the probability [tex]\( P(X = 28) \)[/tex] is approximately [tex]\( 0.1669 \)[/tex].
