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.
+118609 I like your graph Alan. :)
what alan has done is
$$\\ ^nC_1\;\left( \frac{1}{6}\right)^1\;\left(\frac{5}{6}\right)^{n-1}\\\\
$you can take this another step$\\\\
$Say you want your particular number to appear exactly m times, where$\;\; m $Then the probability would be$\\\\
^nC_m\;\left( \frac{1}{6}\right)^m\;\left(\frac{5}{6}\right)^{n-m}\\\\$$
+118609 P (NOT rolling a particular number in n trials) = $$\left(\frac{5}{6}\right)^n$$
so
P(Rolling a particular number in n trials ) = $$1-\left(\frac{5}{6}\right)^n$$
+33616 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:
(The final sentence in the question suggests that Melody's answer is what was being sought.)
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+118609 I like your graph Alan. :)
what alan has done is
$$\\ ^nC_1\;\left( \frac{1}{6}\right)^1\;\left(\frac{5}{6}\right)^{n-1}\\\\
$you can take this another step$\\\\
$Say you want your particular number to appear exactly m times, where$\;\; m $Then the probability would be$\\\\
^nC_m\;\left( \frac{1}{6}\right)^m\;\left(\frac{5}{6}\right)^{n-m}\\\\$$
