Answered

Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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?

Sagot :

shubhamchouhanvrVT

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)  = [tex]^nC_xp^x(1-p)^{n-x}[/tex]

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) = [tex]^{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[/tex]

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) =

[tex]^{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 \\[/tex]

[tex]+^{10}C_4(0.73)^4(0.27)^6[/tex]

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

#SPJ1

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.