To compute the probability of having [tex]\( x \leq 4 \)[/tex] successes in [tex]\( n = 11 \)[/tex] independent trials, with the probability of success in each trial being [tex]\( p = 0.15 \)[/tex], we will use the cumulative binomial probability formula. The binomial probability formula for exactly [tex]\( k \)[/tex] successes in [tex]\( n \)[/tex] trials is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, calculated as [tex]\( \frac{n!}{k!(n-k)!} \)[/tex]
- [tex]\( p \)[/tex] is the probability of success in an individual trial
- [tex]\( (1-p) \)[/tex] is the probability of failure in an individual trial
We need to find the cumulative probability [tex]\( P(X \leq 4) \)[/tex]. This is the sum of the probabilities of getting 0, 1, 2, 3, or 4 successes in 11 trials. Mathematically, it can be expressed as:
[tex]\[ P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \][/tex]
Using the given values [tex]\( n = 11 \)[/tex] and [tex]\( p = 0.15 \)[/tex], the cumulative probability [tex]\( P(X \leq 4) \)[/tex] is:
[tex]\[ P(X \leq 4) \approx 0.9841 \][/tex]
So, the probability of having 4 or fewer successes in 11 trials, with a success probability of 0.15 per trial, rounded to four decimal places, is:
[tex]\[
0.9841
\][/tex]
