V Question 11, 5.3.68 > HW Score: 75.82%, 9.86 of 13 points Points: 0 of 1 Save According to a study published in a reputable science magazine, about 9 women in 100,000 have cervical cancer (C), so P(C) = 0.00009 and P(no C) = 0.99991. The chance that a Pap smear will incorrectly indicate that a woman without cervical cancer has cervical cancer is 0.06. Therefore, P(test pos|no C)=0.06. What is the probability that a randomly chosen woman who has this test will both be free of cervical cancer and test positive for cervical cancer (a false positive)? P(not have C and test positive) = (Round to six decimal places as needed.)