We roll a fair 6-sided die 5 times. What is the probability that exactly 3 of the 5 rolls are either a 1 or a 2?
+1253 There is a 1/3 chance it will be a 1 or 2 and 2/3 chance it won't. So...
(1/3)^3 x (2/3)^2 = 4/243
You are very welcome!
:P
(i was lazy for latex :)
+26367 We roll a fair 6-sided die 5 times.
What is the probability that exactly 3 of the 5 rolls are either a 1 or a 2?
\(\begin{array}{|rcll|} \hline && \dbinom{5}{3}\left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^2 \\\\ &=& \dbinom{5}{3}\left(\dfrac{1}{3}\right)^3\left(\dfrac{2}{3}\right)^2 \\\\ &=& \dfrac{5}{3}\cdot\dfrac{4}{2}\cdot\dfrac{3}{1}\cdot \dfrac{1}{3^3}\cdot \dfrac{4}{3^2} \\\\ &=& 10\cdot \dfrac{4}{3^5} \\\\ &=& 10\cdot \dfrac{4}{243} \\\\ &=& \dfrac{40}{243} \\\\ &=& 0.16460905350\ (16.5 \%) \\ \hline \end{array} \)
