Question 1012410
{{{P(n)=C(10,n)*(1/3)^n*(2/3)^(10-n)}}}
a) When {{{n=0}}}
{{{P(0)=C(10,0)*(1/3)^0*(2/3)^(10-0)}}}
{{{P(0)=1*1*(2/3)^(10)}}}
{{{P(0)=1024/59049}}}
b) When {{{n=1}}}
{{{P(1)=C(10,1)*(1/3)*(2/3)^9}}}
{{{P(1)=5120/59049}}}
c) At least 2 means {{{n>=2}}}, but you know that,
{{{P(n=0)+P(n=1)+P(n>=2)=1}}}
So then,
{{{1024/59049+5120/59049+P(n>=2)=1}}}
{{{P(n>=2)=1-6144/59049}}}
{{{P(n>=2)=52905/59049}}}
{{{P(n>=2)=17635/19683}}}