# Question

According to a study published in Scientific American, about 8 women in 100,000 have cervical cancer (which we’ll call event C), so P(C) = 0.00008. Suppose the chance that a Pap smear will detect cervical cancer when it is present is 0.84. Therefore,

P(test pos | C) = 0.84

What is the probability that a randomly chosen woman who has this test will both have cervical cancer AND test positive for it?

P(test pos | C) = 0.84

What is the probability that a randomly chosen woman who has this test will both have cervical cancer AND test positive for it?

## Answer to relevant Questions

About 8 women in 100,000 have cervical cancer (C), so P(C) = 0.00008 and P(no C) = 0.99992. The chance that a Pap smear will incorrectly indicate that a woman without cervical cancer has cervical cancer is 0.03. ...The table shows the results of rolling a fair six-sided die. Using the table, find the empirical probability of rolling a 1 for 20, 100, and 1000 trials. Report the theoretical probability of rolling a 1 with a fair ...a. Explain how you could use a random number table (or the random numbers generated by software or a calculator) to simulate rolling a fair four-sided die 20 times. Assume you are interested in the probability of rolling a ...Suppose all the days of the week are equally likely as birthdays. Alicia and David are two randomly selected, unrelated people. a. What is the probability that they were both born on Monday? b. What is the probability that ...Eric wants to go skiing tomorrow, but only if there are 3 inches or more of new snow. According to the weather report, any amount of new snow between 1 inch and 6 inches is equally likely. The probability density curve for ...Post your question

0