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November 20th, 2009, 03:18 AM
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| | Joint PMFs of Multiple Random Variables - Urgent Help It says my answers are incorrect. Can anyone help me please? What did I do wrong..?
On a given day, your golf score takes values from range 100 to 109, with probability 0.1, independently from other days. Determined to improve your score, you decide to play on three different days and declare as your score the minimum X of the scores X1, X2, X3 on the different days. - Calculate the PMF of X.
pX(107)=
pX(107)=comb(3,1)*0.1*0.3*0.3=0.027
Px(k)=P(X>k-1)- P(X>k)
Where P(X>k) = P(X1>k, X2>k, X3>k)=(109-k)^3*/(10^3) - pX(101)=
pX(101)= comb(3,1)*0.1*0.9*0.9=0.243; - By how much has your expected score changed as a result of playing on three days?
If PX(100+i)=0.3*(1-i)^2 then E(X)=102.475
If P(X.1 then E(X)=104.5 Difference=102.475-104.5=-2.025 | 
November 20th, 2009, 05:54 PM
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| | Counting To get a score of 107 in the new 'minimum' fashion, he can only get scores of 107, 108 or 109 on the 3 days. The probability of getting a 107 is 0.1, the probability of getting a 108 or 109 is 0.2. His scores could have been:
aaa - 3 scores of 107
aab - 2 scores of 107, and 1 of 108 or 109
abb - 1 scores of 107, and 2 of 108 or 109.
aaa - happens in 1 way
aab - happens in 3 ways.
abb - happens in 3 ways.
aaa - probability = 0.1*0.1*0.1=0.001
aab - probability = 0.1*0.1*0.2=0.002
abb - probability = 0.1*0.2*0.2=0.004
So the total probability for 107 is:
1*0.001 + 3*0.002 + 3*0.004 = 0.019 | | The following users thank qmech for this useful post: | |  | 
November 21st, 2009, 02:26 AM
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| | Quote:
Originally Posted by qmech To get a score of 107 in the new 'minimum' fashion, he can only get scores of 107, 108 or 109 on the 3 days. The probability of getting a 107 is 0.1, the probability of getting a 108 or 109 is 0.2. His scores could have been:
aaa - 3 scores of 107
aab - 2 scores of 107, and 1 of 108 or 109
abb - 1 scores of 107, and 2 of 108 or 109.
aaa - happens in 1 way
aab - happens in 3 ways.
abb - happens in 3 ways.
aaa - probability = 0.1*0.1*0.1=0.001
aab - probability = 0.1*0.1*0.2=0.002
abb - probability = 0.1*0.2*0.2=0.004
So the total probability for 107 is:
1*0.001 + 3*0.002 + 3*0.004 = 0.019 | Thank you...And, how can I compute the probability of 101? | 
November 21st, 2009, 09:28 AM
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| | Method's the same, just different numbers The probability of the score 101 is still 0.1, but the probability of getting a larger score is now different. Larger scores include 102, 103, 104, 105, 106, 107, 108 and 109. There are 8 of them, so that likelihood is 0.8, instead of 0.2 for the last problem.
The method's the same, can you try it with the different numbers? | | The following users thank qmech for this useful post: | |  | | Thread Tools | | | | Display Modes | Linear Mode |
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