Data Model and Decision Making

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|Homework: | DATA MODEL AND DECISION MAKING by Demitri Betrimas |

• Exercise 2.4
• Exercise 2.9
• Exercise 2.24

Exercise 2.4 –
a) What is the probability of finding oil at all three of the sites? The table below shows the different events and there probabilities Event Symbol Probability
The event of finding oil at 1st site A 0.70
The event of finding oil at 2nd site B0.85
The event of finding oil at 3rd site C0.80
Since all three events are independent and does not affect the outcome of the other so the probability of finding oil at all three sites can be known by applying “Implication of fourth law of probability” P (A and B and C) = P (A) * P (B) * P(C)

= 0.70 * 0.85 * 0.80
= 0.476
b) What is the probability of not finding oil at any of the three sites? Probability of NOT finding oil at event A= 1– 0.70 = 0.30 Probability of NOT finding oil at event A= 1– 0.85 = 0.15 Probability of NOT finding oil at event A= 1– 0.80 = 0.20 So Probability of not finding all oil at any sites is

P(Not A and Not B and Not C ) = P(Not A) * P(Not B) * P(Not C) = 0.30 * 0.15 * 0.20
= 0.009
Exercise 2.9 – a) For constructing the probability table let us presume that H – denote the event that person is Infected with HIV
V – denote the event that person is NOT infected with HIV
Y – denote the event that person is Tested positive in the Blood test N – denote the event that person is Tested negative in the Blood test Other data provided to us in the question is
Total people infected with HIV – 550,000 (0.55 mil)
Drug user with HIV infection – 275,000 (0.275 mil)
Remaining with HIV infection – 275,000 (0.275 mil)
Total drug user – 10 mil
Total US population – 250 mil
Probability of any person being infected by HIV P(H) = 0.55/250 = 0.0022 Probability of any person NOT being infected by HIV P(V) = 1 – P (H) = 0.9978 Probability of person infected with HIV and testing positive for HIV P(Y│H) =0.99 Probability of person Not infected with HIV and testing negative for HIV P(N│V) =0.99 We know that

P (Y and H) = P (Y│H) * P (H) = 0.99 * 0.0022 = 0.002178
P (N and H) = 0.0022 – 0.002178 = 0.000022
Also, we know that
P(N and V) = P (N│V) * P (V) = 0.99 * 0.99780 = 0.987822
P (Y and V) = 0.99780 – 0.987822 = 0.99780
Below is the probability table
|  |H |V |Total | |Y |0.0021780 |0.0099780 |0.0121560 | |N |0.0000220 |0.9878220 |0.9878440 | |Total |0.0022000 |0.9978000 |1.0000000 | |  |  |  |  |

We are interested in finding the probability of a person being infected with HIV when the test result is positive i.e. P(H│Y) = P(H and Y) / P(Y) = 0.002178 /0.012156 = 0.17917
Based on the computation there is a 17.9% probability that the person who is tested positive is actually HIV infected. B) Probability of a drug user being infected by HIV is P(H’)= 0.275/10 = 0.0275 Probability of a drug user NOT being infected by HIV is P(V’) =1–P(H’) =0.9725 Probability of person infected with HIV and testing positive for HIV P(Y’│H’) =0.99 Probability of person Not infected with HIV and testing negative for HIV P(N’│V’) =0.99 We know that

P (Y’ and H’) = P (Y’│H’) * P (H’) = 0.99 * 0.0275 = 0.027225 P (N’ and H’) = 0.0022 – 0.002178 = 0.0002750
Also, we know that
P(N’ and V’) = P (N’│V’) * P (V’) = 0.99 * 0.97250 = 0.962775 P (Y’ and V’) = 0.97250 – 0.962775 = 0.0097250
Below is the probability table
|  |H' |V' |Total | |Y' |0.0272250 |0.0097250...
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