A certain medical test is known to detect 72% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test, what is the probability that the test will show that: All 10 have the disease, rounded to four decimal places? At least 8 have the disease, rounded to four decimal places? At most 4 have the disease, rounded to four decimal places?

Answer: 10773Step-by-step explanation:.75^10=0.0563 look at 8, 9, 10 8 the probability is 10C8*0.75^8*0.25^2=0.2816 9 the probability is 0.1877 using the same approach At least 8 is 0.5256   At most 4 is  0 .25^10=essentially 0 1: 10C1*0.75*0.25^9=<0.0001 2: 45*0.75^2*0.25^8=0.0004 3: 120*0.75^3*0.25^7=0.0031 4: 10C4*0.75^4*0.25^6=0.0162 That sum is 0.0197