This is a binomial probability problem because we have a fixed number of trials (10 students), two possible outcomes for each student (either they attend the Tet festivities or they don't), and the probability of success (attending) is constant (25% or 0.25).

The binomial probability formula is:

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

Where:

• $n$ is the number of trials (students surveyed, $n = 10$)
• $k$ is the number of successes (students attending)
• $p$ is the probability of success (probability a student attends, $p = 0.25$)
• $\binom{n}{k}$ is the binomial coefficient, which represents the number of ways to choose $k$ successes from $n$ trials

We are asked to find the probability that at most 3 students attend the festivities, which means we need to calculate the cumulative probability for $X \leq 3$. In other words:

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

Step-by-step calculation:

1. For $P(X = 0)$:

$P(X = 0) = \binom{10}{0} (0.25)^0 (0.75)^{10} = 1 \times 1 \times 0.0563 = 0.0563$

2. For $P(X = 1)$:

$P(X = 1) = \binom{10}{1} (0.25)^1 (0.75)^9 = 10 \times 0.25 \times 0.1937 = 0.4843$

3. For $P(X = 2)$:

$P(X = 2) = \binom{10}{2} (0.25)^2 (0.75)^8 = 45 \times 0.0625 \times 0.2316 = 0.3258$

4. For $P(X = 3)$:

$P(X = 3) = \binom{10}{3} (0.25)^3 (0.75)^7 = 120 \times 0.0156 \times 0.3121 = 0.2920$

Final Calculation:

Now, summing these probabilities:

$P(X \leq 3) = 0.0563 + 0.4843 + 0.3258 + 0.2920 = 1.1584$

Rounding this to four decimal places, the probability is approximately:

$P(X \leq 3) \approx 1.1584$

So, the probability that at most 3 student