You could have also counted them with order not counting as you did. As long as you are consistent.

]]>So you have to count both possibilities.

Bob

]]>Have a look at my table below.

There are 12 possible outcomes so your probabilities should be /12

Bob

]]>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?

]]>*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?

]]>