Given the following data, and rounding to four decimal places: What is the proba

Question

Given the following data, and rounding to four decimal places:

What is the probability of:

an individual being diagnosed with Tuberculosis AND having HIV? Blank 1

randomly selecting an individual from the provided sample who has not been diagnosed with Tuberculosis OR has not been diagnosed with HIV? Blank 2

an individual having HIV given that they have been diagnosed with TB? Blank 3

an individual having HIV given that they have not been diagnosed with TB? Blank 4

Explanation / Answer

Here first the total frequency is computed as: = 249 + 3451 + 2978 + 40,568 = 47246

a) Probability that a person diagnosed with TB has HIV as well:

P( TB+ and HIV+ ) = 249 / 47246 = 0.0053

Therefore 0.0053 is the required probability here.

b) Probability of randomly selecting an individual from the provided sample who has not been diagnosed with Tuberculosis or has not been diagnosed with HIV

= 1 - Probability who has been diagnosed with both

= 1 - 0.0053 = 0.9947

Therefore 0.9947 is the required probability here.

c) Probability of a person having HIV given that they have been diagnosed with TB. Using Bayes theorem, this is computed as:

P( HIV | TB ) = ( Number of people both HIV + and TB + ) / ( Total number of people TB + )

= 249 / (249 + 3451 ) = 0.0673

Therefore 0.0673 is the required probability here.

d) Probability that an individual have HIV given that they dont have TB. Again using Bayes theorem, we get:

P( HIV + | TB -) = ( Number of people with HIV + and TB - ) / ( Total number of people with TB - )

= ( 2978 ) / (2978 + 40568)

= 0.0684

Therefore 0.0684 is the required probability here.

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