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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

 

 

 

 

cheeky

cheifraidhunter  Jun 7, 2017

Best Answer 

 #3
avatar+815 
+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
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3+0 Answers

 #1
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letter C

Guest Jun 7, 2017
 #2
avatar+75333 
+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

 

 

 

cool cool cool

CPhill  Jun 7, 2017
 #3
avatar+815 
+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

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