Acertain medical test is known to detect 56% 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?

The probability that all 10 have the disease is 0.0030

The probability that at least 8 have the disease is 0.1111

The probability that at most 4 have the disease is 0.2407

Step-by-step explanation:

Consider the provided information.

A certain medical test is known to detect 56% of the people who are afflicted with the disease Y. i.e p = 56% = 0.56

Let X be the number number of people who are afflicted with the disease Y.

10 people with the disease are administered the test. n=10

As we know p = 0.56 so q = 1-0.56 = 0.44

Part (A) All 10 have the disease, rounded to four decimal places.

Substitute respective values in the above formula.

The probability that all 10 have the disease is 0.0030

Part (B) At least 8 have the disease.

P(X≥8)=P(X=8)+P(X=9)+P(X=10)

The probability that at least 8 have the disease is 0.1111

Part (C) At most 4 have the disease,  

P(X≤4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

The probability that at most 4 have the disease is 0.2407

answer: haha (: i’ll go see what it is

step-by-step explanation:

thus, the first answer choice, y = sec x, is the correct one