Probability: four 1s, three 2s, two 3s, one 4 in urn

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bbash30

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An urn contains ten numbered balls- four 1's, three 2's, two 3's, and one 4.

a. two balls are drawn without replacement, what is the probability that the sum is 5?

b. two balls are drawn with replacement, what is the probability that the sum is 5?
 
Re: Probability! Please Help!

What have you tried? Where are you having trouble?
Please show evidence of effort.
 
Re: Probability! Please Help!

I am just not sure where to begin. Do we use the Sample Point Approach?
I have paired up numbers to show all the different possibilities, like:
1-1
1-2
1-3
1-4
However, than you have more than one of each number so how do you compute all that?
 
Make an outcomes table:
(1,1), (1,2), (1,3), (1,4)
(2,1), (2,2), (2,3), (2,4)
(3,1), (3,2), (3,3), (3,4)
(4,1), (4,2), (4,3), (4,4)
If we do not have replacement the ‘diagonal’ element (4,4) is impossible.
Thus with replacement:
\(\displaystyle P\left( {(4,4)} \right) = \left( {\frac{1}{{10}}} \right)^2 \,\mbox{but}\,P\left( {(1,1)} \right) = \left( {\frac{4}{{10}}} \right)^2\).
So you need to find the probability of each of those outcomes.
Note a sum of five can only happen with {(1,4), (4,1), (3,2), (2,3)}.
Add up the probabilities.

Then without replacement:
\(\displaystyle P\left( {(4,4)} \right) = 0\,\mbox{but}\,P\left( {(1,1)} \right) = \left( {\frac{4}{{10}}} \right) \left( {\frac{3}{{10}}} \right)\).
 
P(sum of 5)=4/100+6/100+6/100+4/100=20/100= 1/5

Is that correct for the "with replacement"?
 
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