Question

A certain medical test is known to detect 73% 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 1

The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:

• P(At least 8 have the disease) ≈ 0.4378
• P(At most 4 have the disease)  ≈ 0.0342

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as

X = B(n,p)

The probability that out of n trials, there'd be x successes is given by

P(X=x)  = $^nC_xp^x(1-p)^{n-x}$

Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.

Let we say

Success= Probability of a diseased person tagged as diseased by the clinic

Failure = Probability of a diseased person tagged as not diseased by the clinic.

Then,

P(Success) = p = 72% = 0.72 (of a single person)

P(Failure) = q = 1-p = 0.28

Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)

Calculating the needed probabilities, we get:

a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)

P(X≥8) = $^{10}C_8(0.73)^8(0.28)^2+^{10}C_9(0.73)^9(0.27)^1+^{10}C_{10}(0.73)^{10}(0.27)^0$

P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378

b)  P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)

P (X ≤ 4) =

$^{10}C_0(0.73)^0(0.27)^{10}+^{10}{C_1(0.73)^1(0.27)^9+^{10}{C_2(0.73)^2(0.27)^8+^{10}C_3(0.73)&^3(0.27)^7$

$+^{10}C_4(0.73)^4(0.27)^6$

P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719

P (X ≤ 4) =  0.0342

Thus,

The probabilities of considered events are:

• P(At leased 8 have disease) = 0.4378 approx
• P(At most 4 have the disease)  = 0.0342 approx

Learn more about binomial distribution here:

brainly.com/question/13609688

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