Statistics Exercise with Solution

Let's break down the calculation and explanation for \(P(X \leq 3)\) when \(X\) is a binomial random variable with parameters \(n = 10\) (number of trials) and \(p = 0.3\) (probability of success on each trial).

### Binomial Distribution

The binomial distribution describes the probability of obtaining exactly \(k\) successes out of \(n\) trials, with each trial having two possible outcomes: success (with probability \(p\)) or failure (with probability \(1-p\)).

The probability mass function (PMF) for a binomial distribution is given by:

\[ P(X = k) = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} \]

### Calculating \(P(X \leq 3)\)

To find \(P(X \leq 3)\), we sum the probabilities of \(X\) being 0, 1, 2, and 3:

\[ P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \]

Let's calculate each component:

1. **\(P(X=0)\):**

\[ P(X = 0) = \frac{10!}{0!(10-0)!} (0.3)^0 (0.7)^{10} = (0.7)^{10} \]

2. **\(P(X=1)\):**

\[ P(X = 1) = \frac{10!}{1!(10-1)!} (0.3)^1 (0.7)^9 = 10 \times 0.3 \times (0.7)^9 \]

3. **\(P(X=2)\):**

\[ P(X = 2) = \frac{10!}{2!(10-2)!} (0.3)^2 (0.7)^8 = 45 \times (0.3)^2 \times (0.7)^8 \]

4. **\(P(X=3)\):**

\[ P(X = 3) = \frac{10!}{3!(10-3)!} (0.3)^3 (0.7)^7 = 120 \times (0.3)^3 \times (0.7)^7 \]

Summing up these probabilities gives us \(P(X \leq 3)\).

### Conclusion

Using the factorial notation to express binomial coefficients, we've analytically shown how to calculate the cumulative probability \(P(X \leq 3)\) for a binomial random variable \(X\) with parameters \(n = 10\) and \(p = 0.3\), which we previously found to be approximately 0.650. This method sums the probabilities of getting up to 3 successes in 10 trials, demonstrating the application of binomial distribution principles.