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

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

Where:

• $n$ is the number of trials,
• $p$ is the probability of success on a single trial,
• $x$ is the number of successes,
• $\binom{n}{x}$ is the binomial coefficient.

You are asked to find the probability of getting $x \leq 3$ successes when $n = 9$ and $p = 0.5$. This means we need to sum the probabilities for $x = 0, 1, 2, 3$:

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

Let’s calculate each of these probabilities:

For $x = 0$:

$P(X = 0) = \binom{9}{0} \cdot 0.5^0 \cdot (1 - 0.5)^9 = 1 \cdot 1 \cdot 0.001953125 = 0.001953125$

For $x = 1$:

$P(X = 1) = \binom{9}{1} \cdot 0.5^1 \cdot (1 - 0.5)^8 = 9 \cdot 0.5 \cdot 0.00390625 = 0.017578125$

For $x = 2$:

$P(X = 2) = \binom{9}{2} \cdot 0.5^2 \cdot (1 - 0.5)^7 = 36 \cdot 0.25 \cdot 0.0078125 = 0.0703125$

For $x = 3$:

$P(X = 3) = \binom{9}{3} \cdot 0.5^3 \cdot (1 - 0.5)^6 = 84 \cdot 0.125 \cdot 0.015625 = 0.1640625$

Now, sum all these probabilities:

$P(X \leq 3) = 0.001953125 + 0.017578125 + 0.0703125 + 0.1640625 = 0.25390625$

Thus, the probability of $x \leq 3$ successes is approximately:

$\boxed{0.2539}$

Would you like more details or have any questions on this solution?

Here are 5 related questions to deepen your understanding:

1. How do you compute binomial coefficients in the probability formula?
2. What happens to binomial probabilities if $p$ is different from 0.5?
3. How can the binomial probability formula be used for large $n$?
4. What is the cumulative distribution function in binomial probability?
5. How do we approximate binomial probabilities using the normal distribution?

Tip: Always check if a problem asks for "less than or equal to" ($\leq$) or "exactly" for correct interpretation of cumulative probabilities.