Today I am going to share basic concepts for Dice problems from Probability.

### Types

- Cards
- Balls
- Coins
- Dice

### If two dices are thrown: Sum of dices

If 1 appears on the first dice and 1 on the second dice, the sum = 2 (Minimum sum)(1, 1) = 1+1=2

If 1 appears on the first dice and 2 on the second dice, the sum = 3

(1, 2) = 1+2=3

If 1 appears on the first dice and 3 on the second dice, the sum = 4

(1, 3) = 1+3=4

If 1 appears on the first dice and 4 on the second dice, the sum = 5

(1, 4) = 1+4=5

If 1 appears on the first dice and 5 on the second dice, the sum = 6

(1, 5) = 1+5=6

If 1 appears on the first dice and 6 on the second dice, the sum = 7

(1, 6) = 1+6=7

After that dice will repeat like:

(2, 1) = 2+1=3

(2, 2) = 2+2=4

(2, 3) = 2+3=5

(2, 4) = 2+4=6

(2, 5) = 2+5=7

(2, 6) = 2+6=8

Then gain

(3, 1) = 3+1=4

………….

………….

………….

(3, 6) = 3+6=9

Similarly if

(6, 1) = 6+1=7

……….

………..

And so on

(6, 6) = 6+6=12 (Maximum)

Now if two dices are rolled together

Minimum sum =2

Maximum sum = 12

Now if the question says

### What is the probability of getting a sum of 4 if two dices are rolled together?

We can see that the sum of 4 can be obtained by:Solution: Total out comes = 6*6=36 (because two dices are thrown so the total outcome will be 36)

The required probability = Favorable outcomes /Total outcomes = 3/36=1/12

The required probability = Favorable outcomes /Total outcomes = 3/36=1/12

### What is the probability of getting a sum of 6 if two dices are thrown?

**Solution:**

The sum of 6 can be obtained by:

So the probability = 5/36

Note: Memorize this picture to solve probability sums

Solution: From the picture

Sum 4: Probability 3

Sum 12: Probability 1

So the required probability = 3/36+1/36 =4/36=1/9.

**Now to find out the probability of a sum when two dices are thrown, there is a shortcut which is based on the diagram below.**

Let us solve some problems

So the required probability = 4/36= 1/9.

From the picture if the sum is 10, the probability is 3.

So the required probability is 3/36 that is 1/12.

Solution: From the picture

Sum 6: probability 5

Sum 11: probability 2

The required probability = 5/36+2/36= 7/36

**What is the probability of getting a sum of 5 if two dices are thrown together?**

**Solution:**Observe the picture very carefully we can see if the sum is 5, the probability is 4.

So the required probability = 4/36= 1/9.

**What is the probability of getting a sum of 10 if two dices are thrown together?**

From the picture if the sum is 10, the probability is 3.So the required probability is 3/36 that is 1/12.

**What is the probability of getting a sum of 6 or 11 if two dices are rolled together?**

Sum 6: probability 5

Sum 11: probability 2

The required probability = 5/36+2/36= 7/36

**What is the probability of getting a sum of 4 or 12 if two dices are rolled together?**

Solution: From the pictureSum 4: Probability 3

Sum 12: Probability 1

So the required probability = 3/36+1/36 =4/36=1/9.

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