In the game of Twister, a spinner randomly selects either an arm or a leg, and also selects one of four colors, each with equal probability, and players have to move the appropriate body part to the appropriately colored spot on the ground. There are four players. Each player spins once, and makes the move the spinner selects. What is the probability that in these four spins, there will be exactly two moves to a red spot, and the body part selected to move will be an arm exactly $3$ times?
I know that for the red spot, it will be $\dbinom 4 2$ or $6$, $\times (1/4)^2$, $\times (3/4)^2$, which ends up as $6 \times 1/16 \times 9/16$, or $27/128$ when simplified. After this though, how do you factor in the arm?
Also, help with latex would be appreciated to make the numbers bold.
According to your wording (I am not a Twister expert), the spinner's choice of limb and of color are independent, and the players act independently, so
$\Pr[R2 \cap A3] = \Pr[R2] Pr[A3] = (\frac{1}{4})^2 (\frac{3}{4})^2 \binom{4}{2} \times (\frac{1}{2})^4\binom{4}{3} = \frac{54}{1024} = \frac{27}{512}$