Question: Solve QUESTION 1 part b: The probability that a patient recovers from a rare blood disease is 5%. If 15 people are known to have contracted this disease, determine the probability that: (i) there is no survival (ii) at least one survived (iii) exactly one survived (iv) at most one survived.
1. **Problem Statement:**
We have a binomial distribution scenario where the probability of success (patient recovers) is $p=0.05$ and the number of trials (patients) is $n=15$.
2. **Formula:**
The binomial probability formula is:
$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$
where $X$ is the number of successes, $k$ is a specific number of successes.
3. **Calculations:**
(i) Probability of no survival ($k=0$):
$$P(X=0) = \binom{15}{0} (0.05)^0 (0.95)^{15} = 1 \times 1 \times 0.95^{15} = 0.4633$$
(ii) Probability of at least one survived:
$$P(X \geq 1) = 1 - P(X=0) = 1 - 0.4633 = 0.5367$$
(iii) Probability of exactly one survived ($k=1$):
$$P(X=1) = \binom{15}{1} (0.05)^1 (0.95)^{14} = 15 \times 0.05 \times 0.95^{14} = 0.3677$$
(iv) Probability of at most one survived ($k \leq 1$):
$$P(X \leq 1) = P(X=0) + P(X=1) = 0.4633 + 0.3677 = 0.8310$$
4. **Explanation:**
- We use the binomial distribution because each patient either recovers or does not, independently.
- The probability of no survival is the chance all 15 fail to recover.
- At least one survival is the complement of no survival.
- Exactly one survival uses the binomial formula for $k=1$.
- At most one survival sums probabilities for $k=0$ and $k=1$.
**Final answers:**
- (i) $P(X=0) = 0.4633$
- (ii) $P(X \geq 1) = 0.5367$
- (iii) $P(X=1) = 0.3677$
- (iv) $P(X \leq 1) = 0.8310$