jmc

There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to pick which $2$ of the $4$ spins will come up red. Each spin has a $1/4$ chance of coming up red, and $3/4$ chance of not coming up red, so once we pick which $2$ spins we want to come up red, there is a ${\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^2$ chance that the two spins we pick come up red and the other two do not. So, there is a $$6{\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^2=\frac{27}{128}$$chance that exactly $2$ come up red.

There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ ways to pick which 3 of the 4 spins will point to an arm. Each spin has a $1/2$ chance of pointing to an arm, and $1/2$ chance of not pointing to an arm, so once we pick which $3$ spins we want to point to an arm, there is a ${\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^1$ chance that the three spins we pick come up arm and the other one does not. So, there is a $$4{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^1=\frac{1}{4}$$chance that exactly $3$ spins point to an arm.

The color selection and limb selection are independent events, so the probability that they both happen is the product of their individual probabilities; $\frac{27}{128}\cdot \frac{1}{4}=\boxed{\frac{27}{512}}$.