$P(3)$ Probability of exactly 3 successes: 0.336415625

### Solution:

$P(3)$ Probability of exactly 3 successes

If using a calculator, you can enter $\text{trials} = 5$, $p = 0.65$, and $X = 3$ into a binomial probability distribution function (PDF). If doing this by hand, apply the binomial probability formula: $$P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X}$$ The binomial coefficient, $\binom{n}{X}$ is defined by $$\binom{n}{X} = \frac{n!}{X!(n-X)!}$$ The full binomial probability formula with the binomial coefficient is $$P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X}$$ where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $n = 5$, $p = 0.65$, and $X = 3$. $$P(3) = \frac{5!}{3!(5-3)!} \cdot 0.65^3 \cdot (1-0.65)^{5-3}$$ Evaluating the expression, we have $$P(3) = 0.336415625$$

### Complete Binomial Distribution Table

If we apply the binomial probability formula, or a calculator's binomial probability distribution (PDF) function, to all possible values of X for 5 trials, we can construct a complete binomial distribution table. The sum of the probabilities in this table will always be 1. The complete binomial distribution table for this problem, with p = 0.65 and 5 trials is:

P(0) = 0.0052521875
P(1) = 0.0487703125
P(2) = 0.181146875
P(3) = 0.336415625
P(4) = 0.3123859375
P(5) = 0.1160290625