To find the probability of obtaining [tex]\( x \leq 3 \)[/tex] successes in a series of [tex]\( n = 9 \)[/tex] independent trials with a success probability [tex]\( p = 0.6 \)[/tex] for each trial, we need to use a binomial probability distribution.
The binomial probability formula for exactly [tex]\( k \)[/tex] successes in [tex]\( n \)[/tex] trials is:
[tex]\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\][/tex]
where [tex]\(\binom{n}{k}\)[/tex] represents the binomial coefficient, calculated as:
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
To find [tex]\(P(X \leq 3)\)[/tex], we sum up 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]
However, rather than calculating each term individually, we often refer to cumulative distribution functions (CDFs) which give us the direct probability of [tex]\( X \)[/tex] being less than or equal to a certain value.
Given your parameters ([tex]\( n = 9 \)[/tex], [tex]\( p = 0.6 \)[/tex]), and the required cumulative probability ([tex]\( x \leq 3 \)[/tex]), the cumulative probability [tex]\( P(X \leq 3) \)[/tex] calculates to:
[tex]\[
P(X \leq 3) = 0.09935257600000003
\][/tex]
Thus, rounding this to four decimal places, we get:
[tex]\[
P(X \leq 3) \approx 0.0994
\][/tex]
So, the probability of having [tex]\( x \leq 3 \)[/tex] successes in 9 trials with a success rate of 0.6 is approximately [tex]\( 0.0994 \)[/tex].
