To find the probability of [tex]\( x \)[/tex] successes where [tex]\( x \leq 3 \)[/tex] in [tex]\( n = 9 \)[/tex] independent trials with a success probability [tex]\( p = 0.2 \)[/tex] for each trial, we will use the binomial probability formula.
The binomial probability formula 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]\[ \binom{n}{k} = \frac{n!}{k! (n - k)!} \][/tex]
Let us calculate [tex]\( P(X = k) \)[/tex] for [tex]\( k = 0, 1, 2, 3 \)[/tex] and then sum these probabilities:
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[
P(X = 0) = \binom{9}{0} (0.2)^0 (0.8)^9 = 1 \cdot 1 \cdot 0.134217728 \approx 0.1342
\][/tex]
So, [tex]\( P(X = 0) \approx 0.1342 \)[/tex].
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[
P(X = 1) = \binom{9}{1} (0.2)^1 (0.8)^8 = 9 \cdot 0.2 \cdot 0.16777216 \approx 0.3020
\][/tex]
So, [tex]\( P(X = 1) \approx 0.3020 \)[/tex].
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[
P(X = 2) = \binom{9}{2} (0.2)^2 (0.8)^7 = 36 \cdot 0.04 \cdot 0.2097152 \approx 0.3020
\][/tex]
So, [tex]\( P(X = 2) \approx 0.3020 \)[/tex].
4. For [tex]\( x = 3 \)[/tex]:
[tex]\[
P(X = 3) = \binom{9}{3} (0.2)^3 (0.8)^6 = 84 \cdot 0.008 \cdot 0.262144 \approx 0.1762
\][/tex]
So, [tex]\( P(X = 3) \approx 0.1762 \)[/tex].
Now, we sum these probabilities to find the total probability of [tex]\( x \leq 3 \)[/tex]:
[tex]\[
P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
\][/tex]
[tex]\[
P(X \leq 3) \approx 0.1342 + 0.3020 + 0.3020 + 0.1762 = 0.9144
\][/tex]
Therefore, the probability of obtaining [tex]\( x \leq 3 \)[/tex] successes in 9 independent trials with a success probability of 0.2 is approximately [tex]\( 0.9144 \)[/tex].
