A die is thrown twice. What is the probability that the sum of the number of dots is 4 or 10?

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Since the total number of faces of a die is 6, the probability to get one of the six faces of the die when it is thrown is:

P = 1/6

We know that this die is thrown twice, therefore the probability becomes:

P = 1/6*6 = 1/36

Now, let's analyze what combination of faces leads to the sum of number of dots of 4 or 10.

The value of the sum of dots is 4 when we have the following combinations:

1 + 3 = 4

2 + 2 = 4

3 + 1 = 4

Keeping in mind that the total number of dots a face can have is 6, the value of the sum of dots is 10 when we have the following combinations:

4 + 6 = 10

5 + 5 = 10

6 + 4 = 10

Therefore, we've identified 6 posssible outcomes, therefore the probability is:

P = 6/36

P = 1/6

**Therefore, the probability that the sum of the number of dots is 4 or 10 is P = 1/6.**

When a die is thrown twice, there are 36 possible outcomes.

3 of these outcomes give a sum of 4:

- 1 and 3
- 2 and 2
- 3 and 1

3 of these outcomes give a sum of 10:

- 4 and 6
- 5 and 5
- 6 and 4

Therefore, there are 6 outcomes that produce a sum of 4 or 10.

The probability is 6/36 = 1/6

We will create the following table to sort out the problem

FIRST DICE -------->

1| 2| 3| 4| 5| 6

--------------------------------------------

1 | 2| 3| 4| 5| 6| 7

2 | 3| 4| 5| 6| 7| 8

3 | 4| 5| 6| 7| 8| 9

4 | 5| 6| 7| 8| 9| 10

5 | 6| 7| 8| 9| 10| 11

6 | 7| 8| 9| 10| 11| 12

From this table we can get the sum of the two dots in both cases.

thus, all possible outcome = 36

favourable outcome = 3 + 3 =6

probability = 6/36 **= 1/6(Ans.)**

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