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?

- Thread starter bbash30
- Start date

a. two balls are drawn

b. two balls are drawn

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?

- Joined
- Jan 29, 2005

- Messages
- 7,829

(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)

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:

\(\displaystyle P\left( {(4,4)} \right) = 0\,\mbox{but}\,P\left( {(1,1)} \right) = \left( {\frac{4}{{10}}} \right) \left( {\frac{3}{{10}}} \right)\).