To solve this problem, we need to calculate the cumulative binomial probability of having [tex]$x \leq 3$[/tex] successes in 9 independent trials with the probability of success [tex]$p = 0.7$[/tex] in each trial.
The binomial probability formula for having 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, representing the number of ways to choose [tex]$k$[/tex] successes from [tex]$n$[/tex] trials, calculated as [tex]\( \frac{n!}{k!(n-k)!} \)[/tex].
- [tex]\( p^k \)[/tex] is the probability of having [tex]$k$[/tex] successes.
- [tex]\( (1-p)^{n-k} \)[/tex] is the probability of having [tex]$n-k$[/tex] failures.
To find the cumulative probability [tex]\( P(X \leq 3) \)[/tex], we sum the probabilities of having 0, 1, 2, and 3 successes.
Step-by-step, we calculate each probability:
1. For [tex]\( k = 0 \)[/tex] successes:
[tex]\[
P(X = 0) = \binom{9}{0} (0.7)^0 (0.3)^9
\][/tex]
2. For [tex]\( k = 1 \)[/tex] success:
[tex]\[
P(X = 1) = \binom{9}{1} (0.7)^1 (0.3)^8
\][/tex]
3. For [tex]\( k = 2 \)[/tex] successes:
[tex]\[
P(X = 2) = \binom{9}{2} (0.7)^2 (0.3)^7
\][/tex]
4. For [tex]\( k = 3 \)[/tex] successes:
[tex]\[
P(X = 3) = \binom{9}{3} (0.7)^3 (0.3)^6
\][/tex]
Finally, sum up these individual probabilities:
[tex]\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\][/tex]
After performing the computation for each term and summing them, the cumulative probability comes out to be:
[tex]\[
P(X \leq 3) = 0.025294842
\][/tex]
Therefore, the probability of having 3 or fewer successes out of 9 independent trials, with each trial having a success probability of 0.7, is approximately [tex]\(0.0253\)[/tex].
