Answer:
Using law of total probability.
Probability a person diagnosed with cancer is 0.05+0.78+0.95+0.06=0.096
Answer:
The answer is 0.406
Step-by-step explanation:
Exercise 2.95: In a certain region of the country it is known from past experience that theprobability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and theprobability of incorrectly diagnosing a person without cancer as having the disease is 0.06,what is the probability that an adult over 40 years of age is diagnosed as having cancer?
This time we want to calculate P(C | D). To do this, we use Bayes rule:
[tex]P(C | D) = \frac{P(C n D)}{P(D)} \\= \frac{P(C)P(D | C)}{P(C)P(D | C) + P(C`)P(D | C`)} \\= \frac{0.05 * 0.78}{(0.05 * 0.78) + (0.95 * 0.06)} \\= \frac{0.039}{0.039 + 0.057} \\=\frac{0.039}{0.096} \\= 0.406[/tex]
