1. **Problem statement:** Calculate the probability that in 80 independent roulette spins, the ball lands on a specific number at least 4 times. 2. **Model and formula:** The number of times the ball lands on a specific number in 80 spins follows a binomial distribution $X \sim \text{Binomial}(n=80, p=\frac{1}{37})$ because each number has an equal probability $p=\frac{1}{37}$. 3. **We want:** $$P(X \geq 4) = 1 - P(X \leq 3) = 1 - \sum_{k=0}^3 \binom{80}{k} \left(\frac{1}{37}\right)^k \left(1-\frac{1}{37}\right)^{80-k}$$ 4. **Calculate each term:** - $P(X=0) = \binom{80}{0} \left(\frac{1}{37}\right)^0 \left(\frac{36}{37}\right)^{80} = \left(\frac{36}{37}\right)^{80}$ - $P(X=1) = \binom{80}{1} \left(\frac{1}{37}\right)^1 \left(\frac{36}{37}\right)^{79} = 80 \cdot \frac{1}{37} \cdot \left(\frac{36}{37}\right)^{79}$ - $P(X=2) = \binom{80}{2} \left(\frac{1}{37}\right)^2 \left(\frac{36}{37}\right)^{78} = \frac{80 \cdot 79}{2} \cdot \left(\frac{1}{37}\right)^2 \cdot \left(\frac{36}{37}\right)^{78}$ - $P(X=3) = \binom{80}{3} \left(\frac{1}{37}\right)^3 \left(\frac{36}{37}\right)^{77} = \frac{80 \cdot 79 \cdot 78}{6} \cdot \left(\frac{1}{37}\right)^3 \cdot \left(\frac{36}{37}\right)^{77}$ 5. **Sum these probabilities:** $$P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$$ 6. **Final probability:** $$P(X \geq 4) = 1 - P(X \leq 3)$$ This gives the probability that the ball lands on the chosen number at least 4 times in 80 spins. **Note:** For exact numerical value, use a calculator or software to evaluate the sums.