To solve this problem, we will be using the concept of cumulative binomial probability, where we are interested in the probability of getting at most \( x \) successes in \( n \) independent trials with a success probability \( p \).
Given:
- Number of trials (\( n \)) = 9
- Probability of success (\( p \)) in a single trial = 0.2
- Number of successes (\( x \leq 3 \)).
The formula for the binomial probability of exactly \( k \) successes in \( n \) trials is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \][/tex]
Where \( \binom{n}{k} \) is the binomial coefficient calculated as:
[tex]\[ \binom{n}{k} = \frac{n!}{k! (n - k)!} \][/tex]
To find the cumulative probability \( P(X \leq 3) \), we need to sum the probabilities of getting 0, 1, 2, and 3 successes:
[tex]\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
Calculating each term:
1. For \( k = 0 \):
[tex]\[ P(X = 0) = \binom{9}{0} (0.2)^0 (0.8)^9 \][/tex]
2. For \( k = 1 \):
[tex]\[ P(X = 1) = \binom{9}{1} (0.2)^1 (0.8)^8 \][/tex]
3. For \( k = 2 \):
[tex]\[ P(X = 2) = \binom{9}{2} (0.2)^2 (0.8)^7 \][/tex]
4. For \( k = 3 \):
[tex]\[ P(X = 3) = \binom{9}{3} (0.2)^3 (0.8)^6 \][/tex]
Summing these, we get the cumulative probability:
[tex]\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \][/tex]
Using these calculations (or equivalent steps) and rounding the cumulative probability to four decimal places, we get:
The probability of [tex]\( x \leq 3 \)[/tex] successes is approximately [tex]\( 0.9144 \)[/tex].
