**a regular 6 sided die is rolled three times. find the probability of each event:**

**A) Three 6s in a row**

**B) 5,1,even**

**C) Odd, greater than 2, 5**

cheifraidhunter
Jun 7, 2017

#3**+2 **

Of course, a dice has 6 faces that are all equally likely to occur.

A) Let's make a table that illustrates the probability of just one 6 showing up on one roll:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✘ | ||

6 | ✓ | ||

Probability | -------------------- | 1/6 | 5/6 |

As you can see from this table, there is only a 1/6 probability of getting a 6 the first time. What about three times in a row? Multiply the probabilities of all independent events together. Each event has a 1/6 chance.

\(P=(\frac{1}{6})^3=\frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{1}{216}\approx0.46\%\)

Therefore, there is a 1/216 chance in rolling a 6 three times in a row.

B)

We'll use the same method as above. However, these events aren't all the same, so we'll have to calculate the events independently.

Let's do it for rolling a 5

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✓ | ||

6 | ✘ | ||

Probability | -------------------- | 1/6 | 5/6 |

The probability of rolling a five is alos 1/6. Let's try the next event for seeing the chances of rolling a 1:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✓ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✘ | ||

6 | ✘ | ||

Probability | -------------------- | 1/6 | 5/6 |

Here, too, there is a 1/6 probability of rolling a 1. Let's find the possibility of rolling an even number:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✓ | ||

3 | ✘ | ||

4 | ✓ | ||

5 | ✘ | ||

6 | ✓ | ||

Probability | -------------------- | 3/6=1/2 | 3/6=1/2 |

Now that we have determined the probability of each event occurring. Multiply the probabilities of each event together to get the total probability:

\(\frac{1}{6}*\frac{1}{6}*\frac{1}{2}=\frac{1}{72}\approx1.39\%\)

Therefore, there is a 1/72 chance of rolling a 5, then a 1, and then an even number.

C)

Use the same method as illustrated above.

Let's first find the probability of getting an odd number:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✓ | ||

2 | ✘ | ||

3 | ✓ | ||

4 | ✘ | ||

5 | ✓ | ||

6 | ✘ | ||

Probability | -------------------- | 3/6=1/2 | 3/6=1/2 |

Now, let's find the probability of a number greater than 2:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✓ | ||

4 | ✓ | ||

5 | ✓ | ||

6 | ✓ | ||

Probability | -------------------- | 4/6=2/3 | 2/6=1/3 |

Finally, the last one, which is rolling a 5:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✓ | ||

6 | ✘ | ||

Probability | -------------------- | 1/6 | 5/6 |

Multiply the probabilities

\(\frac{1}{2}*\frac{2}{3}*\frac{1}{6}=\frac{2}{36}=\frac{1}{18}\approx5.56\%\)

YOu're done now!

TheXSquaredFactor
Jun 7, 2017

#2**+1 **

A) Three 6's = (1/6)^3 = 1 / 216

B) 5, 1, even = (1/6) (1/6) (1/2) = 1 / 72

C) Odd, >2, 5 = (1/2) (4 / 6) (1/6) = ( 1/2) (2/3) (1/6) 2 / 36 = 1 / 18

CPhill
Jun 7, 2017

#3**+2 **

Best Answer

Of course, a dice has 6 faces that are all equally likely to occur.

A) Let's make a table that illustrates the probability of just one 6 showing up on one roll:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✘ | ||

6 | ✓ | ||

Probability | -------------------- | 1/6 | 5/6 |

As you can see from this table, there is only a 1/6 probability of getting a 6 the first time. What about three times in a row? Multiply the probabilities of all independent events together. Each event has a 1/6 chance.

\(P=(\frac{1}{6})^3=\frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{1}{216}\approx0.46\%\)

Therefore, there is a 1/216 chance in rolling a 6 three times in a row.

B)

We'll use the same method as above. However, these events aren't all the same, so we'll have to calculate the events independently.

Let's do it for rolling a 5

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✓ | ||

6 | ✘ | ||

Probability | -------------------- | 1/6 | 5/6 |

The probability of rolling a five is alos 1/6. Let's try the next event for seeing the chances of rolling a 1:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✓ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✘ | ||

6 | ✘ | ||

Probability | -------------------- | 1/6 | 5/6 |

Here, too, there is a 1/6 probability of rolling a 1. Let's find the possibility of rolling an even number:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✓ | ||

3 | ✘ | ||

4 | ✓ | ||

5 | ✘ | ||

6 | ✓ | ||

Probability | -------------------- | 3/6=1/2 | 3/6=1/2 |

Now that we have determined the probability of each event occurring. Multiply the probabilities of each event together to get the total probability:

\(\frac{1}{6}*\frac{1}{6}*\frac{1}{2}=\frac{1}{72}\approx1.39\%\)

Therefore, there is a 1/72 chance of rolling a 5, then a 1, and then an even number.

C)

Use the same method as illustrated above.

Let's first find the probability of getting an odd number:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✓ | ||

2 | ✘ | ||

3 | ✓ | ||

4 | ✘ | ||

5 | ✓ | ||

6 | ✘ | ||

Probability | -------------------- | 3/6=1/2 | 3/6=1/2 |

Now, let's find the probability of a number greater than 2:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✓ | ||

4 | ✓ | ||

5 | ✓ | ||

6 | ✓ | ||

Probability | -------------------- | 4/6=2/3 | 2/6=1/3 |

Finally, the last one, which is rolling a 5:

Possible Events | Meets Condition | Does Not Meet Condition | |

1 | ✘ | ||

2 | ✘ | ||

3 | ✘ | ||

4 | ✘ | ||

5 | ✓ | ||

6 | ✘ | ||

Probability | -------------------- | 1/6 | 5/6 |

Multiply the probabilities

\(\frac{1}{2}*\frac{2}{3}*\frac{1}{6}=\frac{2}{36}=\frac{1}{18}\approx5.56\%\)

YOu're done now!

TheXSquaredFactor
Jun 7, 2017