Sleep Disorder Probability 4Bd693

1. **State the problem:** We want to find the probability that at least 3 people out of 25 have a sleep disorder, given that the probability any one person has a sleep disorder is $p=0.21$. 2. **Identify the distribution:** This is a binomial probability problem where $n=25$ and $p=0.21$. 3. **Formula:** The binomial probability for exactly $k$ successes is $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ 4. **Calculate $P(X \geq 3)$:** It is easier to use the complement rule: $$P(X \geq 3) = 1 - P(X < 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$ 5. **Calculate each term:** $$P(X=0) = \binom{25}{0} (0.21)^0 (0.79)^{25} = 1 \times 1 \times 0.79^{25} = 0.79^{25}$$ $$P(X=1) = \binom{25}{1} (0.21)^1 (0.79)^{24} = 25 \times 0.21 \times 0.79^{24}$$ $$P(X=2) = \binom{25}{2} (0.21)^2 (0.79)^{23} = 300 \times 0.21^2 \times 0.79^{23}$$ 6. **Calculate powers and values:** $$0.79^{25} \approx 0.0081$$ $$0.79^{24} \approx 0.0103$$ $$0.79^{23} \approx 0.0130$$ 7. **Calculate probabilities:** $$P(X=0) = 0.0081$$ $$P(X=1) = 25 \times 0.21 \times 0.0103 = 25 \times 0.21 \times 0.0103 = 0.0541$$ $$P(X=2) = 300 \times 0.0441 \times 0.0130 = 300 \times 0.0441 \times 0.0130 = 0.1720$$ 8. **Sum and subtract from 1:** $$P(X<3) = 0.0081 + 0.0541 + 0.1720 = 0.2342$$ $$P(X \geq 3) = 1 - 0.2342 = 0.7658$$ **Final answer:** $$\boxed{0.766}$$ (rounded to three decimal places)