A die is rolled 18 times and two threes come up. Find the probability of this event.
+118723 Your logic looks good to me Chris
nCr(18,2)*(1/6)^2*(5/6)^16
$${\left({\frac{{\mathtt{18}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{18}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{{\mathtt{16}}} = {\mathtt{0.229\: \!873\: \!544\: \!962\: \!680\: \!4}}$$
Nearly 23% just like you said :)))
+3502 isnt it just 1/6 no matter how many times you roll it since there are only six sides to a dye well considering its a six sided one
+130511 Check me out here, Melody.....
We can choose any two of the 18 rolls to have the 3 occur.......and the probability that it occurs on any one roll is (1/6)......and the probaility that it doesn't occur is just (5/6)
So the probaility is
C(18, 2)* (1/6)^2 * (5/6)^16 = about 23%
+118723 Your logic looks good to me Chris
nCr(18,2)*(1/6)^2*(5/6)^16
$${\left({\frac{{\mathtt{18}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{18}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}^{{\mathtt{16}}} = {\mathtt{0.229\: \!873\: \!544\: \!962\: \!680\: \!4}}$$
Nearly 23% just like you said :)))
