Question

The probability of a randomly selected adult in one country being infected with a certain virus is 0.003. In tests for the virus, blood samples

from 29 people are combined. What is the probability that the combined sample tests positive for the virus? Is it unlikely for such a combined

sample to test positive? Note that the combined sample tests positive if at least one person has the virus.

The probability that the combined sample will test positive is

(Round to three decimal places as needed.)

Answer 1

Answer:

The probability that the combined sample tests positive  for the virus is 0.083

Since the probability that combined sample test positive for the virus is greater than 0.05, it is not likely for such a combined  sample to test positive.

The probability that the combined sample will test positive is 0.083

Step-by-step explanation:

Given that:

The probability of a randomly selected adult in one country being infected with a certain virus is 0.003.

P = 0.003

number of blood sample size n = 29

The probability mass function of X is as follows;

$P(X=x) = \left[\begin{array}{c}{29}&x\\\end{array}\right] (0.003)^x (1-0.003)^{29-x}$

Thus; the required probability is;

$P(X \geq 1) = 1 - P ( X < 1)$

$P(X \geq 1) = 1 - P ( X =0)$

$P(X \geq 1) = 1 - \left[\begin{array}{c} \dfrac{29!}{0!(29-0)!} \ \ \times 0.003)^0 \times (1-0.003)^{29-0}}\end{array}\right]$

$P(X \geq 1) = 1 - \left[\begin{array}{c} 1 \times 1 \times ( 0.9166)\end{array}\right]$

$P(X \geq 1) = 1 - 0.9166$

$P(X \geq 1) = 0.0834$

Therefore; the probability that the combined sample tests positive  for the virus is 0.083

Is it unlikely for such a combined  sample to test positive?

P(combined  sample test positive  for the virus ) = 0.0834

Since the probability that combined sample test positive for the virus is greater than 0.05, it is not likely for such a combined  sample to test positive.

The probability of a randomly selected adult in one country being infected with a certain virus is 0.003.

P = 0.003

number of blood sample size n = 29

The probability mass function of X is as follows;

$P(X=x) = \left[\begin{array}{c}{29}&x\\\end{array}\right] (0.003)^x (1-0.003)^{29-x}$

Thus; the required probability is;

$P(X \geq 1) = 1 - P ( X < 1)$

$P(X \geq 1) = 1 - P ( X =0)$

$P(X \geq 1) = 1 - \left[\begin{array}{c} \dfrac{29!}{0!(29-0)!} \ \ \times 0.003)^0 \times (1-0.003)^{29-0}}\end{array}\right]$

$P(X \geq 1) = 1 - \left[\begin{array}{c} 1 \times 1 \times ( 0.9166)\end{array}\right]$

$P(X \geq 1) = 1 - 0.9166$

$P(X \geq 1) = 0.0834$

The probability that the combined sample will test positive is 0.083