To calculate the probability of having at most 3 successes ([tex]\(x \leq 3\)[/tex]) in 9 independent trials of a binomial experiment with the probability of success in each trial [tex]\( p = 0.7 \)[/tex], we can use the binomial probability formula:
[tex]\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \][/tex]
where:
- [tex]\( \binom{n}{x} \)[/tex] is the binomial coefficient
- [tex]\( n \)[/tex] is the number of trials (9 in this case)
- [tex]\( x \)[/tex] is the number of successes
- [tex]\( p \)[/tex] is the probability of success (0.7 in this case)
- [tex]\( (1-p) \)[/tex] is the probability of failure (0.3 in this case)
We need to find the cumulative probability for [tex]\( x \)[/tex] ranging from 0 to 3.
### Step-by-Step Solution:
1. Calculate [tex]\( P(X = 0) \)[/tex]:
[tex]\[
P(X = 0) = \binom{9}{0} (0.7)^0 (0.3)^9 = 1 \times 1 \times 0.000019683 = 0.000019683
\][/tex]
2. Calculate [tex]\( P(X = 1) \)[/tex]:
[tex]\[
P(X = 1) = \binom{9}{1} (0.7)^1 (0.3)^8 = 9 \times 0.7 \times 0.000282429 = 9 \times 0.0001977 = 0.00177963
\][/tex]
3. Calculate [tex]\( P(X = 2) \)[/tex]:
[tex]\[
P(X = 2) = \binom{9}{2} (0.7)^2 (0.3)^7 = 36 \times 0.49 \times 0.002187 = 36 \times 0.00107163 = 0.03857916
\][/tex]
4. Calculate [tex]\( P(X = 3) \)[/tex]:
[tex]\[
P(X = 3) = \binom{9}{3} (0.7)^3 (0.3)^6 = 84 \times 0.343 \times 0.0102 = 84 \times 0.0034986 = 0.2949432
\][/tex]
5. Sum the probabilities:
[tex]\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\][/tex]
[tex]\[
P(X \leq 3) = 0.000019683 + 0.00177963 + 0.03857916 + 0.2849432 = 0.0253
\][/tex]
Therefore, the probability of obtaining at most 3 successes in 9 trials is [tex]\( \boxed{0.0253} \)[/tex].
