1. **Problem Statement:** We have a drug with a survival rate of 70% (probability of survival $p=0.7$) and death rate $q=1-p=0.3$. For 12 women taking the drug, we want to find: A) Probability all 12 die B) Probability exactly 10 survive C) Probability none die (all survive) D) Probability more than 8 survive 2. **Formula Used:** This is a binomial probability problem. The binomial probability formula is: $$P(X = x) = \binom{n}{x} p^x q^{n-x}$$ where $n$ is the number of trials, $x$ is the number of successes (survivals), $p$ is the probability of success, and $q$ is the probability of failure. 3. **Calculations:** A) All die means 0 survive, so $x=0$: $$P(X=0) = \binom{12}{0} (0.7)^0 (0.3)^{12} = 1 \times 1 \times 0.3^{12} = 0.3^{12}$$ Calculate $0.3^{12}$: $$0.3^{12} \approx 5.31441 \times 10^{-7}$$ B) Exactly 10 survive means $x=10$: $$P(X=10) = \binom{12}{10} (0.7)^{10} (0.3)^2$$ Calculate binomial coefficient: $$\binom{12}{10} = \frac{12!}{10!2!} = 66$$ Calculate powers: $$(0.7)^{10} \approx 0.0282475, \quad (0.3)^2 = 0.09$$ Multiply all: $$66 \times 0.0282475 \times 0.09 \approx 0.1677$$ C) None die means all survive, $x=12$: $$P(X=12) = \binom{12}{12} (0.7)^{12} (0.3)^0 = 1 \times 0.7^{12} \times 1 = 0.7^{12}$$ Calculate $0.7^{12}$: $$0.7^{12} \approx 0.01384$$ D) More than 8 survive means $x > 8$, i.e. $x=9,10,11,12$: $$P(X>8) = P(X=9) + P(X=10) + P(X=11) + P(X=12)$$ Calculate each term: - $P(X=9) = \binom{12}{9} (0.7)^9 (0.3)^3$ $$\binom{12}{9} = 220$$ $$(0.7)^9 \approx 0.0403536, (0.3)^3 = 0.027$$ $$P(X=9) = 220 \times 0.0403536 \times 0.027 \approx 0.2397$$ - $P(X=10)$ calculated above as $\approx 0.1677$ - $P(X=11) = \binom{12}{11} (0.7)^{11} (0.3)^1$ $$\binom{12}{11} = 12$$ $$(0.7)^{11} \approx 0.01977, (0.3)^1 = 0.3$$ $$P(X=11) = 12 \times 0.01977 \times 0.3 \approx 0.0712$$ - $P(X=12)$ calculated above as $\approx 0.01384$ Sum all: $$P(X>8) \approx 0.2397 + 0.1677 + 0.0712 + 0.01384 = 0.4924$$ 4. **Final Answers:** - A) $P(\text{all die}) \approx 5.31 \times 10^{-7}$ - B) $P(10 \text{ survive}) \approx 0.168$ - C) $P(\text{none die}) \approx 0.0138$ - D) $P(\text{more than 8 survive}) \approx 0.492$ These probabilities show the likelihood of each event based on the binomial distribution with survival probability 0.7 and 12 trials.