Mathematics, 17.07.2020 14:01 wowlol448

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.)

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Answer from: justijust500

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

Answer from: Quest

step-by-step explanation:

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The dot plot below shows the hourly rate of some babysitters: babysitting rates 51.00 52.00 53.00 5

Answer from: Quest

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

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