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VijaKhara@gmail.com science forum beginner
Joined: 30 Sep 2005
Posts: 26
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Posted: Fri Sep 30, 2005 5:54 pm Post subject:
Can I use conditional probability to solve this problem???
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Dear all,
I always get difficulties when distinguishing conditional probability
of an "intersection" probability to solve a problem.
For example.
a person tosses a fair die and then draw a card from a desk which
contains only number cards (1-->10)
Compute the probability that the card drawn has a numerical value at
least twice that on the face of the die.
I interpret the problem in two ways and solve as follows:
Sol1:
Denoting:
After tossing the die:
A1: toss number 1,
A2:toss number 2
A3: toss a 3.
A4: toss a 4.
A5: toss a 5.
A6: toss a 6.
B1: draw: any of 2,3,4,5,6,7,8,9,10 P(B1)=36/40
B2: draw: any of 4,5,6,7,8,9,10, P(B2)=28/40
B3: draw: any of 6,7,8,9,10, P(B3)=20/40
B4: draw: any of 8,9,10 P(B4)=12/40
B5: draw: 10 P(B5)=4/40
The Event = (A1 and B1)or(A2 and B2)or(A3 and B3)or(A4 and B4)
or(A5 and B5).
Thus,
P=P(A1)*P(B1)+ P(A2)*P(B2)+P(A3)*P(B3)+P(A4)*P(B4)+P(A5)*P(B5)
Solution 2:
But What if I compute and interpret the prob as follows:
Denoting B as the event that the card drawn has a numerical value at
least twice that on the face of the die.
P=P(B|A1)P(A1)+P(B|A2)P(A2)+P(B|A3)P(A3)+P(B|A4)P(A4)+P(B|A5)P(A5)+
P(B|A6)P(A6) where (Ai: toss a number i)
We have
P(B|A1)= 36/40 (equal to P(B1) in the first solution).
P(B|A2)= 28/40 (equal to P(B2) in the first solution).
P(B|A3)= 20/40 (equal to P(B3) in the first solution).
P(B|A4)= 12/40 (equal to P(B4) in the first solution).
P(B|A5)= 1/40 (equal to P(B5) in the first solution).
P(B|A6)=0
If we do this way, the result will be different from Solution 1:
What way is correct? or no one correct? plz help me
Thanks
Vijay
Prob=P |
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VijaKhara@gmail.com science forum beginner
Joined: 30 Sep 2005
Posts: 26
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Posted: Fri Sep 30, 2005 6:01 pm Post subject:
Re: Can I use conditional probability to solve this problem???
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I mean I always get difficult when choosing between conditional
probability or an "intersection" probability method to solve a problem.
...blah blah
Sorry for making confusion.
VijaKh...@gmail.com wrote:
| Quote: | Dear all,
I always get difficulties when distinguishing conditional probability
of an "intersection" probability to solve a problem.
For example.
a person tosses a fair die and then draw a card from a desk which
contains only number cards (1-->10)
Compute the probability that the card drawn has a numerical value at
least twice that on the face of the die.
I interpret the problem in two ways and solve as follows:
Sol1:
Denoting:
After tossing the die:
A1: toss number 1,
A2:toss number 2
A3: toss a 3.
A4: toss a 4.
A5: toss a 5.
A6: toss a 6.
B1: draw: any of 2,3,4,5,6,7,8,9,10 P(B1)=36/40
B2: draw: any of 4,5,6,7,8,9,10, P(B2)=28/40
B3: draw: any of 6,7,8,9,10, P(B3)=20/40
B4: draw: any of 8,9,10 P(B4)=12/40
B5: draw: 10 P(B5)=4/40
The Event = (A1 and B1)or(A2 and B2)or(A3 and B3)or(A4 and B4)
or(A5 and B5).
Thus,
P=P(A1)*P(B1)+ P(A2)*P(B2)+P(A3)*P(B3)+P(A4)*P(B4)+P(A5)*P(B5)
Solution 2:
But What if I compute and interpret the prob as follows:
Denoting B as the event that the card drawn has a numerical value at
least twice that on the face of the die.
P=P(B|A1)P(A1)+P(B|A2)P(A2)+P(B|A3)P(A3)+P(B|A4)P(A4)+P(B|A5)P(A5)+
P(B|A6)P(A6) where (Ai: toss a number i)
We have
P(B|A1)= 36/40 (equal to P(B1) in the first solution).
P(B|A2)= 28/40 (equal to P(B2) in the first solution).
P(B|A3)= 20/40 (equal to P(B3) in the first solution).
P(B|A4)= 12/40 (equal to P(B4) in the first solution).
P(B|A5)= 1/40 (equal to P(B5) in the first solution).
P(B|A6)=0
If we do this way, the result will be different from Solution 1:
What way is correct? or no one correct? plz help me
Thanks
Vijay
Prob=P |
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iandjmsmith@aol.com science forum beginner
Joined: 13 Sep 2005
Posts: 15
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Posted: Sat Oct 01, 2005 6:41 am Post subject:
Re: Can I use conditional probability to solve this problem???
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VijaKhara@gmail.com wrote:
| Quote: | Dear all,
I always get difficulties when distinguishing conditional probability
of an "intersection" probability to solve a problem.
For example.
a person tosses a fair die and then draw a card from a desk which
contains only number cards (1-->10)
Compute the probability that the card drawn has a numerical value at
least twice that on the face of the die.
I interpret the problem in two ways and solve as follows:
Sol1:
Denoting:
After tossing the die:
A1: toss number 1,
A2:toss number 2
A3: toss a 3.
A4: toss a 4.
A5: toss a 5.
A6: toss a 6.
B1: draw: any of 2,3,4,5,6,7,8,9,10 P(B1)=36/40
B2: draw: any of 4,5,6,7,8,9,10, P(B2)=28/40
B3: draw: any of 6,7,8,9,10, P(B3)=20/40
B4: draw: any of 8,9,10 P(B4)=12/40
B5: draw: 10 P(B5)=4/40
The Event = (A1 and B1)or(A2 and B2)or(A3 and B3)or(A4 and B4)
or(A5 and B5).
Thus,
P=P(A1)*P(B1)+ P(A2)*P(B2)+P(A3)*P(B3)+P(A4)*P(B4)+P(A5)*P(B5)
Solution 2:
But What if I compute and interpret the prob as follows:
Denoting B as the event that the card drawn has a numerical value at
least twice that on the face of the die.
P=P(B|A1)P(A1)+P(B|A2)P(A2)+P(B|A3)P(A3)+P(B|A4)P(A4)+P(B|A5)P(A5)+
P(B|A6)P(A6) where (Ai: toss a number i)
We have
P(B|A1)= 36/40 (equal to P(B1) in the first solution).
P(B|A2)= 28/40 (equal to P(B2) in the first solution).
P(B|A3)= 20/40 (equal to P(B3) in the first solution).
P(B|A4)= 12/40 (equal to P(B4) in the first solution).
P(B|A5)= 1/40 (equal to P(B5) in the first solution).
P(B|A6)=0
If we do this way, the result will be different from Solution 1:
What way is correct? or no one correct? plz help me
Thanks
Vijay
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The only difference between the 2 is in the first you have P(B5) = 4/40
and in the second you have P(B|A5) = 1/40, the second being incorrect.
In the first events Bi are B|Ai and so there is no real difference.
Ian Smith |
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