SOLUTION: Many e-mail account holders get spam e-mail inviting the purchase of certain “erectile assistance” drugs like Cialis and Tadalfil. (I get these regularly). Such drugs have to

Question 1166441: Many e-mail account holders get spam e-mail inviting the purchase of certain “erectile assistance” drugs like Cialis and Tadalfil. (I get these regularly). Such drugs have to be approved by the FDA in USA before sale. Suppose there is a 0.95 chance the FDA will approve a new drug if the results of current testing show no side effects; but they will also approve the drug with a 0.50 chance if testing shows side effects. A certain physician working for the drug manufacturer believes there is a 0.20 chance the drug will cause side effects. What is the probability the drug will be approved by the FDA?
Please help, thanks

Answer by CPhill(2030) (Show Source): You can put this solution on YOUR website!
The probability that the new drug will be approved by the FDA is **0.86** (or **86%**).
This problem is solved using the **Law of Total Probability**, which accounts for all possible scenarios leading to the drug's approval.
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## 🎲 Calculation using the Law of Total Probability
Let $A$ be the event that the drug is **Approved**.
Let $S$ be the event that the drug causes **Side Effects**.
Let $S'$ be the event that the drug causes **No Side Effects**.
### 1. Identify Probabilities
| Event | Notation | Probability |
| :---: | :---: | :---: |
| Drug causes side effects | $P(S)$ | $0.20$ |
| Drug causes no side effects | $P(S') = 1 - P(S)$ | $1 - 0.20 = 0.80$ |
| Approved *given* no side effects | $P(A|S')$ | $0.95$ |
| Approved *given* side effects | $P(A|S)$ | $0.50$ |
### 2. Apply the Law of Total Probability
The total probability of approval, $P(A)$, is the sum of the probabilities of being approved in the "no side effects" scenario and the "side effects" scenario:
$$P(A) = P(A|S') \cdot P(S') + P(A|S) \cdot P(S)$$
### 3. Substitute and Compute
$$P(A) = (0.95) \cdot (0.80) + (0.50) \cdot (0.20)$$
$$P(A) = 0.76 + 0.10$$
$$P(A) = \mathbf{0.86}$$
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