# Probability with pairs of things

1. Aug 31, 2007

### mutzy188

1. The problem statement, all variables and given/known data

The five numbers 1, 2, 3, 4, and 5 are written respectively on five disks of the same size and placed in a hat. Two disks are drawn without replacement from the hat, and the numbers written on them is observed.

(a) List the 10 possible outcomes for this experiment as unordered pairs of numbers.

(b) If each of the 10 outcomes has probability 1/10, assign a value to the probability that the sum of the two numbers drawn is (i) 3, (ii) between 6 and 8 inclusive

3. The attempt at a solution

(a)
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5

(b)

(i)

That would be the pair: 1 2

The probability of drawing a 1 and a 2 is: 1/10

(ii)

That would be the pairs:

1 5
2 4
2 5
3 4
3 5
4 5

This probability would be: (1/10)^6

Thanks

2. Aug 31, 2007

### neutrino

Check the your second answer in b.

3. Aug 31, 2007

### mutzy188

How would I go about finding the probability for that one

4. Aug 31, 2007

### wbclark

It is only to correct to multiply probabilities if you're trying to find the probability of ALL of these things happen. In other words, if you needed to draw first the pair 1 5, then 2 4, then 2 5, etc.

However, the condition is that ONE of these combinations is drawn in ONE trial, so you add the probabilities.

5. Sep 1, 2007

### HallsofIvy

Staff Emeritus
(1/10)6 is the probability that you drew those specific pairs in 6 consectutive drawings (with replacement).

The probability of drawing one of those is the number of such outcomes, 5, divided by the total number of outcomes, 10. Of course, that is exactly the same as adding the probability of each, as wbclark said.

Last edited: Sep 1, 2007
6. Sep 1, 2007

### neutrino

To add to that, 4 and 5 make 9.

7. Sep 1, 2007

### HallsofIvy

Staff Emeritus
Picky, picky!

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