In a binomial experiment, the probability of exactly $x$ successes in $n$ trials is given by the binomial probability formula:

$P(X = x) = \binom{n}{x} p^x (1 - p)^{n-x}$

Given:

• $n = 10$ (number of trials)
• $x = 3$ (desired number of successes)
• $p = 0.45$ (probability of success)
• $q = 1 - p = 0.55$ (probability of failure)

Step 1: Compute the binomial coefficient

$\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Step 2: Compute the probability

$P(X = 3) = 120 \times (0.45)^3 \times (0.55)^7$

Let's calculate this in the next step. Given values from math import comb

n = 10 x = 3 p = 0.45 q = 1 - p

Compute binomial probability

binomial_coeff = comb(n, x) probability = binomial_coeff * (px) * (q(n-x))

Round to four decimal places

round(probability, 4)The probability of observing exactly 3 successes in 10 trials is 0.1665 (rounded to four decimal places).

Would you like a step-by-step breakdown of the calculations?

Here are five related questions to explore further:

1. What is the probability of observing at most 3 successes ($P(X \leq 3)$)?
2. What is the probability of observing at least 3 successes ($P(X \geq 3)$)?
3. What is the expected number of successes in this experiment?
4. What is the standard deviation of the number of successes?
5. How does changing $p$ to 0.55 affect the probability of getting exactly 3 successes?

Tip:

The expected value of a binomial distribution is given by $E(X) = np$, and the variance is $\sigma^2 = np(1-p)$.