Styles Part 2: Probability Directions: Perform the following calculations. Pleas
ID: 3049921 • Letter: S
Question
Styles Part 2: Probability Directions: Perform the following calculations. Please show all work You roll a six-sided dice four times. What is the probability of rolling 1, 2, 3, and 4? 1. 2. Answer the following questions using the information below Blood Type O: A: B: AB: 4% 46% 40% 10% Rh FactorWorldwide HIV/AIDS Prevalence Positive: 84% HIV positive: 0.9% Negative: 16% HIV negative: 99.1% a) How many different outcomes are possible combining the above three data? b) What is the probability of having blood type O or blood type B? c) What is the probability of being HIV positive with blood type AB, Rh negative? 3. You have a bag of 10 red, 10 blue, and 30 white marbles a You draw five marbles without returning any to the bag What is the probability of drawing 1 red, 1 red, 1 red, 1 blue, and then white?Explanation / Answer
1)
The required outcomes of rolling a 1,2,3 and 4 :
1234
1243
1324
1342
1423
1432
2134
2314
2341 and so on.
These form 4! = 24 outcomes.
Therefore, matching outcomes = 24 outcomes
Total outcomes = 6 x 6 x 6 x 6 = 1296 outcomes
Ans: Prob of rolling a 1,2,3 and 4 = 24/1296 = 0.0185
2.
a) The number of different outcomes possible =
4 different blood groups x 2 different Rh factor types x 2 HIV prevalence
= 4 x 2 x 2 = 16
Ans: 16 different outcomes are possible combining the above data.
b) P(blood type = O or blood type = B) = P(blood type = O) + P(blood type = B)
P(Blood type = O) = 46% = 0.46
P(Blood type = B) = 10% = 0.10
P(blood type = O or blood type = B) = 0.46 + 0.10
Ans: P(blood type = O or blood type = B) = 0.56 = 56%
c)
Considering that HIV status, Blood Type and Rh Factor are 3 seperate, independent events, the probability of eahc of them are independent as well.
----> P(HIV positive & Blood type AB & Rh negative) = P(HIV positive) x P(Blood type AB) x P(Rh negative)
P(HIV positive & Blood type AB & Rh negative) = 0.009 x 0.04 x 0.16
Ans: P(HIV positive & Blood type AB & Rh negative) = 0.0000576
3.
a) P( drawing 1 red, 1 red, 1 red, 1 blue, 1 white without replacement) = 10/50 x 9/49 x 8/48 x 10/47 x 30/46
P( drawing 1 red, 1 red, 1 red, 1 blue, 1 white) = 216,000 / 254,251,200
Ans: P( drawing 1 red, 1 red, 1 red, 1 blue, 1 white) = 8.49 x 10-4 = 0.000849
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