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bbash30
04-08-2008, 01:37 PM
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?

pka
04-08-2008, 02:37 PM
What have you tried? Where are you having trouble?
Please show evidence of effort.

bbash30
04-08-2008, 03:24 PM
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?

pka
04-08-2008, 06:31 PM
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:
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:
P\left( {(4,4)} \right) = 0\,\mbox{but}\,P\left( {(1,1)} \right) = \left( {\frac{4}{{10}}} \right) \left( {\frac{3}{{10}}} \right).

bbash30
04-08-2008, 06:50 PM
P(sum of 5)=4/100+6/100+6/100+4/100=20/100= 1/5

Is that correct for the "with replacement"?

pka
04-08-2008, 07:49 PM
Yes that is correct. Good job.