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How do you calculate the chance of rolling a certain number on a six sided dice once (assuming the rolls are completely random) in a certain number of rolls? Ex: rolling a dice 6 times won't get you a 100% chance of rolling a six.

Guest Feb 19, 2015

Best Answer 

 #3
avatar+91469 
+10

I like your graph Alan.   :)

 

what alan has done is  

 

$$\\ ^nC_1\;\left( \frac{1}{6}\right)^1\;\left(\frac{5}{6}\right)^{n-1}\\\\
$you can take this another step$\\\\
$Say you want your particular number to appear exactly m times, where$\;\; m $Then the probability would be$\\\\
^nC_m\;\left( \frac{1}{6}\right)^m\;\left(\frac{5}{6}\right)^{n-m}\\\\$$

 

Melody  Feb 19, 2015
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3+0 Answers

 #1
avatar+91469 
+5

P (NOT rolling a particular number in n trials) = $$\left(\frac{5}{6}\right)^n$$

 

so

 

P(Rolling a particular number in n trials ) =      $$1-\left(\frac{5}{6}\right)^n$$

Melody  Feb 19, 2015
 #2
avatar+26406 
+5

Melody's answer assumes the probability is of the number appearing at least once.

To calculate the probability of it appearing exactly once, use (n/6)(5/6)n-1.

 

The two probabilities are compared as a function of n in the graph below:

 Probabilities:

 

(The final sentence in the question suggests that Melody's answer is what was being sought.)

.

Alan  Feb 19, 2015
 #3
avatar+91469 
+10
Best Answer

I like your graph Alan.   :)

 

what alan has done is  

 

$$\\ ^nC_1\;\left( \frac{1}{6}\right)^1\;\left(\frac{5}{6}\right)^{n-1}\\\\
$you can take this another step$\\\\
$Say you want your particular number to appear exactly m times, where$\;\; m $Then the probability would be$\\\\
^nC_m\;\left( \frac{1}{6}\right)^m\;\left(\frac{5}{6}\right)^{n-m}\\\\$$

 

Melody  Feb 19, 2015

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