If you had two six sided dice and you tossed them, what would be the probability that the sum is 5 or less?

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- March 31st 2007, 10:30 PMgretchenTwo dice
If you had two six sided dice and you tossed them, what would be the probability that the sum is 5 or less?

- March 31st 2007, 11:51 PMJhevon
i'm quite sure there's a neat little formula for this, but i don't remember it, so let's do it the long and hard way.

we want the sum to be at least 5, so it can be 5,4,3 or 2

how many ways are there to get a sum of 5?

1) Dice1 = 4**AND**Dice2 = 1, Probability = (1/6)*(1/6) = 1/36

2) Dice1 = 3**AND**Dice2 = 2, Probability = (1/6)*(1/6) = 1/36

3) Dice1 = 2**AND**Dice2 = 3, Probability = (1/6)*(1/6) = 1/36

4) Dice1 = 1**AND**Dice2 = 4, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is: 1/36 + 1/36 + 1/36 + 1/36 = 4/36 = 1/9

how many ways are there to get a sum of 4?

1) Dice1 = 3**AND**Dice2 = 1, Probability = (1/6)*(1/6) = 1/36

2) Dice1 = 2**AND**Dice2 = 2, Probability = (1/6)*(1/6) = 1/36

3) Dice1 = 1**AND**Dice2 = 3, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is: 1/36 + 1/36 + 1/36 = 3/36 = 1/12

how many ways are there to get a sum of 3?

1) Dice1 = 2**AND**Dice2 = 1, Probability = (1/6)*(1/6) = 1/36

2) Dice1 = 1**AND**Dice2 = 2, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is: 1/36 + 1/36 = 2/36 = 1/18

how many ways are there to get a sum of 2?

1) Dice1 = 1**AND**Dice2 = 1, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is 1/36

So the probability of getting a sum of 5**OR**4**OR**3**OR**2 is:

1/9 + 1/12 + 1/18 + 1/36 = 5/18

so, the probability that the sum is 5 or less is**5/18** - April 1st 2007, 04:25 AMSoroban
Hello, Gretchen!

Quote:

If you had two six-sided dice and you tossed them,

what would be the probability that the sum is 5 or less?

If you're just starting Probability, you might get used to dice problems.

With two dice, there are 36 possible outcomes.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

You don't have to write them out every time.

Now that you've seen them, you can*visualize*them, right?

Now pick out and count the ones you want.

"Five or less": (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)

There are__ten__of them.

Therefore: P(5 or less) .= .10/36 .= .5/18

- April 1st 2007, 06:41 AMJhevon