Answer
$P(X\leq2)=0.046656+0.186624=0.23328$
Work Step by Step
The solution is based on the binomial distribution:
$P(X=x)=C(n,x)*p^{x}*q^{n-x}$
If $x\leq1$, this means that x is either 0 or 1, therefore we have to add the probabilities of these $x$ values.
If $p=0.4$, $n=6$, then $q=1-0.4=0.6$
By substituting into the given variables, we get:
$P(X=0)=C(6,0)*0.4^0*0.6^6=\frac{6!}{0!\times(6-0)!}*0.4^0*0.6^6=0.046656$
$P(X=1)=C(6,1)*0.4^1*0.6^5=\frac{6!}{1!\times(6-1)!}*0.4^1*0.6^5=0.186624$
The sum of these values gives us:
$P(X\leq2)=0.046656+0.186624=0.23328$
