Medical Laboratory Receives 32 Blood Specimens To Check

A Medical Laboratory Receives 32 Blood Specimens To Check For Hiv Ten

A medical laboratory receives 32 blood specimens to check for HIV. Ten specimens actually contain HIV. A worker is accidentally exposed to five specimens.

(a) What is the probability that none of the five specimens contained HIV?

(b) What is the probability that fewer than three of the five specimens contained HIV?

(c) What is the probability that at least two of the five specimens contained HIV?

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Paper For Above instruction

The scenario described involves calculating probabilities related to the presence of HIV in blood specimens, utilizing combinatorial probability methods. This situation can be modeled using the hypergeometric distribution because it involves sequentially selecting a subset of specimens (five) from a finite population (32) without replacement, where there are a specific number of "successes" (HIV-positive specimens, which total 10). This approach accurately reflects the discrete, finite nature of the sampling process and the lack of replacement.

The hypergeometric distribution models the probability of having exactly k successes (HIV-positive specimens) in n draws (specimens exposed to the worker), given a population of size N with K successes. The formula is:

P(X = k) = [C(K, k) * C(N - K, n - k)] / C(N, n)

where C(a, b) is the binomial coefficient "a choose b."

Part (a): Probability that none of the five specimens contained HIV

Here, N = 32, K = 10, n = 5, and we want P(X = 0). Using the formula:

P(X=0) = [C(10, 0) * C(22, 5)] / C(32, 5)

Calculating each binomial coefficient:

  • C(10, 0) = 1
  • C(22, 5) = 26,334
  • C(32, 5) = 201,376

Therefore:

P(X=0) = (1 * 26,334) / 201,376 ≈ 0.1308

Rounded to four decimal places, the probability that none of the five specimens contained HIV is approximately 0.1308.

Part (b): Probability that fewer than three specimens contained HIV

This includes the cases where exactly 0, 1, or 2 specimens contain HIV:

P(X

Calculations:

  • P(1):
C(10, 1) = 10


C(22, 4) = 7,484


P(1) = [10 * 7,484] / 201,376 ≈ 74,840 / 201,376 ≈ 0.3717
  • P(2):
    • C(10, 2) = 45
    

    C(22, 3) = 1,540
    

    P(2) = [45 * 1,540] / 201,376 ≈ 69,300 / 201,376 ≈ 0.3438

    Adding these probabilities:

    P(X

    Rounded to four decimal places, the probability that fewer than three specimens contained HIV is approximately 0.8463.

    Part (c): Probability that at least two specimens contained HIV

    This is the complement of the probability that fewer than two specimens contain HIV:

    P(X≥2) = 1 - P(X

    Using previous calculations:

    P(0) = 0.1308
    

    P(1) = 0.3717
    

    P(X

    Therefore:

    P(X≥2) = 1 - 0.5025 = 0.4975

    Rounded to four decimal places, the probability that at least two specimens contained HIV is approximately 0.4975.

    Conclusion

    Through applying the hypergeometric distribution, we have found that the probability that no specimens contained HIV is approximately 0.1308, that fewer than three contained HIV is approximately 0.8463, and that at least two contained HIV is approximately 0.4975. These calculations highlight the significance of understanding probability distributions in the context of finite populations, particularly in medical and health-related environments, where such statistical measures are crucial for risk assessment and resource allocation.

    References

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