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Hi, I am a little confused about this problem:
An urn contains four balls numbered 1,2,3 and 4. If two balls are drawn from the urn at random (that is, each pair has the same chance if being selected) and Z is the sum of the numbers on the two balls drawn, find:
a) the probability distribution of Z
What I got for the answer was:
Urn= {1,2,3,4}
Z= {3,4,5,6,7}
p(3) = 1/4 //(1 way to get 3)
p(4) = 1/4 //(1 way to get 4)
p(5) = 1/2 //(2 ways to get 5)
p(6) = 1/4 //(1 way to get 6)
p(7) = 1/4 //(1 way to get 7)
Is this was the question was asking for? I wasn't quite sure. Though I think my answer is wrong because aren't they supposed to add up to 1?
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Hi;
Though I think my answer is wrong because aren't they supposed to add up to 1?
I would like them very much to add up to one.
I get
2 ways to make a 3. ( 1 , 2 )and ( 2 , 1 ).
2 ways to make a 4. ( 1 , 3 )and ( 3 , 1 ).
4 ways to make a 5. ( 1 , 4 ),( 4, 1 ) and ( 2 , 3 )( 3 , 2 )
2 ways to make a 6. ( 2 , 4 )and ( 4 , 2 ).
2 ways to make a 7 ( 3 , 4 )and ( 4 , 3 ).
Can you finish now?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi genericname,
Have a look at my table below.
There are 12 possible outcomes so your probabilities should be /12
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Oh, didn't know you can count them twice like that. Thank you for the replies.
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It's because you could draw a 1 followed by a 2 and also draw a 2 followed by a 1.
So you have to count both possibilities.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi genericname;
You could have also counted them with order not counting as you did. As long as you are consistent.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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