Answer
$P(1\leq X\leq3)=0.186624+0.31104+0.27648=0.1792$
Work Step by Step
The solution is based on the binomial distribution:
$P(X=x)=C(n,x)*p^{x}*q^{n-x}$
If $1\leq x\leq3$, this means that x is either 1, 2 or 3, 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:
$P(X=1)=C(6,1)*0.4^1*0.6^5=\frac{6!}{1!\times(6-1)!}*0.4^1*0.6^5=0.186624$
$P(X=2)=C(6,2)*0.4^2*0.6^4=\frac{6!}{2!\times(6-2)!}*0.4^2*0.6^4=0.31104$
$P(X=3)=C(6,3)*0.4^3*0.6^3=\frac{6!}{3!\times(6-3)!}*0.4^3*0.6^3=0.27648$
The sum of these values gives us:
$P(1\leq X\leq3)=0.186624+0.31104+0.27648=0.1792$
