Question 1032170
{{{p(x) = (mu^x/x!)*e^(-mu) = (3.2^x/x!)*e^(-3.2)}}}.

The probability in question is {{{p(0) + p(1) = (3.2^0/0!)*e^(-3.2)+(3.2^1/1!)*e^(-3.2) = 0.1712}}} to four significant figures.