According to the Centers for Disease Control and Prevention, 1 in every 10,000 pre-screened volunteer blood donors in the United States is infected with the Human Immunodeficiency Virus, or HIV. To reduce the chance of transmitting this virus to other individuals, each unit of donated blood is subject to an immunoassay test before the blood is used for transfusion. This test yields one of two possible outcomes: positive, which indicates that the blood donor has antibodies indicating the presence of HIV, or negative, which indicates that the donor does not have antibodies indicating the presence of HIV. The standard immunoassay test is not 100% accurate. On average, the test erroneously returns a negative test result for 1 in every 100 donors that are, in fact, infected with HIV. Moreover, the test erroneously returns a positive test result for 1 in every 1000 donors that are, in fact, not infected with HIV. Disclaimer: The numbers here are somewhat dated. (Note: Because of the small probabilities involved in this problem, carry out all intermediate calculations to at least seven decimal places or, better yet, do all calculations in Excel or Stata.) (a) What is the probability that a unit of blood from a randomly selected pre-screened volunteer blood donor in the U. S. will test positive? (b) A unit of blood is tested, and the test result is positive. Given this test result, what is the probability that the donor does not actually have HIV?

(a) P(test+) = 0.0010989

(b) P(no HIV | test+) = 0.9099

Step-by-step explanation:

The probability of HIV in blood donors is 1 in every 10000, i.e. P(HIV) = 1/10000 = 0.0001

Let us denote a negative test result with test- and a positive test result with test+. The probability that the test results are negative given that the donor is infected with HIV is 1 in every 100 i.e. P(test- | HIV) = 1/100 = 0.01.

The test returns a positive result for donors who are not infected with HIV with a probability of 1/1000 i.e. P(test+ | no HIV) = 0.001.

We can compute the following probabilities:

P(no HIV) = 1 - P(HIV)

                = 1 - 0.0001

P(no HIV) = 0.9999

P(test+ | HIV) = 1 - P(test- | HIV)

                      = 1 - 0.01

P(test+ | HIV) = 0.99

(a) Now, we need to compute the probability that the donor's blood will test positive i.e. P(test+).

P(test+) = P(test+ ∩ HIV) + P(test+ ∩ no HIV)

            = P(test+ | HIV)*P(HIV) + P(test+ | no HIV)*P(no HIV)

            = (0.99)(0.0001) + (0.001)(0.9999)

            = 0.000099 + 0.0009999

P(test+) = 0.0010989

(b) Now, we need to compute the probability that the donor does not have HIV given that the test result is positive i.e. P(no HIV | test+). Using the conditional probability formula:

P(no HIV | test+) = P(no HIV ∩ test+) / P(test+)

                           = P(test+ | no HIV)*P(no HIV)/ P(test+)

We have found the value of P(test+) in part (a) so,

P(no HIV | test+) = (0.001*0.9999)/0.0010989

                          = 0.0009999/0.0010989

P(no HIV | test+) = 0.9099

(23*2)-12=x     x=34

step-by-step explanation:

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d