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1. Two dice are thrown once at the same time. a. Prepare a table showing the number of possible outcomes. b. Find the probability of getting a total of: i, at least 5 ii. 10 iii. a total score divisible by 5.​

User Tdy
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Answer:

Explanation:

a. The number of possible outcomes when two dice are thrown can be represented in a 6x6 table, as each die has 6 faces. Here's the table:

| Die 1 \ Die 2 | 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|

b. The probability of an event is calculated as the number of favorable outcomes divided by the total number of outcomes.

i. For getting a total of at least `5`, the favorable outcomes are from `5` to `12`. There are `4` ways to get `5`, `5` ways to get `6`, `6` ways to get `7`, `5` ways to get `8`, `4` ways to get `9`, `3` ways to get `10`, `2` ways to get `11`, and `1` way to get `12`. So, there are a total of `30` favorable outcomes. The total number of outcomes is `36` (since each die has `6` faces, and two dice are thrown). So, the probability is $$\frac{30}{36} = \frac{5}{6}$$.

ii. For getting a total of exactly `10`, there are only three possibilities: `(4,6)`, `(5,5)`, and `(6,4)`. So, the probability is $$\frac{3}{36} = \frac{1}{12}$$.

iii. For getting a total score divisible by `5`, the possible totals are `5` and `10`. As calculated above, there are `4` ways to get a total of `5` and `3` ways to get a total of `10`. So, there are a total of `7` favorable outcomes. So, the probability is $$\frac{7}{36}$$.

User Rarblack
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