# Joint PMFs of Multiple Random Variables - Urgent Help

• Nov 20th 2009, 02:18 AM
essedra
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.

1. 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)
2. pX(101)=

pX(101)= comb(3,1)*0.1*0.9*0.9=0.243;
3. 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
• Nov 20th 2009, 04:54 PM
qmech
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
• Nov 21st 2009, 01:26 AM
essedra
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?
• Nov 21st 2009, 08:28 AM
qmech
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?