A binomial experiment has the given number of trials $n$ and the given success probability $p$. \[ x=10, p=03 \] Part: $0 / 3$ Part 1 of 3 (a) Desermine the probablity $P(3)$. Round the anwwer to at leatt three decimal places. \[ P(3)=6936^{\circ} \]
Solution
Step 1 ：We are given a binomial experiment with 10 trials and a success probability of 0.3. We are asked to find the probability of getting exactly 3 successes.
Step 2 ：The formula for the probability mass function of a binomial distribution is: \(P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success on each trial, and \(\binom{n}{k}\) is the binomial coefficient, which is the number of ways to choose \(k\) successes out of \(n\) trials.
Step 3 ：In this case, we have \(n=10\), \(k=3\), and \(p=0.3\). So we need to calculate: \(P(3) = \binom{10}{3} \cdot 0.3^3 \cdot (1-0.3)^{10-3}\).
Step 4 ：Calculating the binomial coefficient \(\binom{10}{3}\), we get 120.
Step 5 ：Substituting these values into the formula, we get a probability of 0.267.
Step 6 ：Final Answer: The probability \(P(3)\) is \(\boxed{0.267}\).
