Question 1124810: A certain medical test is known to detect 75% 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 by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! .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
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