Question 373867
Look at all of the possible outcomes for incorrect(I) and correct(C) placement of labels on each of the 4 boxes.
{{{CCCC}}}
{{{CCCI}}}
{{{CCIC}}}
{{{CCII}}}
{{{CICC}}}
{{{CICI}}}
{{{CIIC}}}
{{{CIII}}}
{{{ICCC}}}
{{{ICCI}}}
{{{ICIC}}}
{{{ICII}}}
{{{IICC}}}
{{{IICI}}}
{{{IIIC}}}
{{{IIII}}}
There are {{{4^2=16}}} possible outcomes.
{{{1}}} has exactly {{{4}}} correct.
{{{4}}} have exactly {{{3}}} correct.
{{{6}}} have exactly {{{2}}} correct.
{{{4}}} have exactly {{{1}}} correct.
{{{1}}} has exactly {{{0}}} correct.
.
.
The probability that the label is correct is {{{P(C)=1/4}}}.
The probability that the label is incorrect is {{{P(I)=3/4}}}.
.
.

{{{P(3C)=4*(1/4)(1/4)(1/4)(3/4)=3/64}}}
{{{P(2C)=6*(1/4)(1/4)(3/4)(3/4)=27/128}}}
{{{P(1C)=4*(1/4)(3/4)(3/4)(3/4)=27/64}}}
At least one correct means that its not the case that all four are incorrect.
{{{P(4I)=1*(3/4)(3/4)(3/4)(3/4)=81/256}}}
So then either all 4 are incorrect or at least one is correct.
P(at least 1 C)+P(4I)={{{1}}}
P(at least 1 C)={{{1-P(4I)}}}
P(at least 1 C)={{{1-81/256}}}
P(at least 1 C)={{{175/256}}}