Neuroblastoma is a rare, serious, but treatable disease. A urine test, the VMA test, has been developed that gives a positive diagnosis in about 70% of cases of neuroblastoma. It has been proposed that this test be used for large-scale screening of children. Assume that 300,000 children are to be tested, of whom 8 have the disease. We are interested in whether or not the test detects the disease in the 8 children who have the disease. Find the probability that a) none will be missed. b) seven cases will be detected. c) two or more cases will be missed. d) exactly 75% will be detected. d) six or less cases will be detected.

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akiran007 akiran007 · Answer 1 · 2020-09-30T06:43:28+02:00

Answer:

Part a) probability that none will be missed means none will have the disease= 0.058

Part b) probability that seven cases will be detected= 0.00122

Part c)probability that two or more cases will be missed.=0.744

Part d) ) probability that exactly 75% will be detected= 75% of 8= 6= 0.01000188

Part e) probability six or less cases will be detected = 0.171

Step-by-step explanation:

This is a binomial probability distribution . Here p= 0.7 and q= 1-0.7= 0.3 n= 8

Part a) none will be missed means none will have the disease

P (x= 0)= 8C0(0.7)^8 (0.3)^0

P (x= 0)=1*(0.7)^8 (0.3)^0=0.058

Part b) seven cases will be detected

P (x= 7)= 8C7(0.7)^1 (0.3)^7 = 0.00122

Part c) two or more cases will be missed.

1- P (x= 0)- P (x= 1)

1-0.058-0.198=0.744

Part d) ) exactly 75% will be detected= 75% of 8= 6

P (x= 6)= 8C6(0.7)^2 (0.3)^6=0.01000188

Part e) six or less cases will be detected.

1- P (x= 7)= 8C7(0.7)^1 (0.3)^7+P (x= 8)= 8C8(0.7)^0 (0.3)^8

1- 0.802+0.027=1- 0.829= 0.171