Die 1 has 4 red and 2 blue faces.
Die 2 has 2 red & 4 blue faces.
A fair coin is flipped once. If it lands on heads, the game continues with Die 1; if it lands on tails, then Die 2 is used.
A) If the first 2 throws result in red, what is the probability of red on the third throw?
B) If red turns up on the first 2 throws, what is the probability that Die 1 is being used?
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I honestly have no clue how to do this problem, please help.
P(Head) = $\frac{1}{2}$, P(Tail) = $\frac{1}{2}$
P(Red with Die 1) = $P(R_1)$ = $\frac{4}{6}$
P(Red with Die 2) = $P(R_2)$ = $\frac{2}{6}$
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A) P($\frac{RRR}{RR}$) = $\frac{P(H)P(R_1)^3+P(T)P(R_2)^3}{P(H)P(R_1)^2+P(T)P(R_2)^2}$
= $\frac{3}{5}$
Best Answer
Probability of using die1 is 0.5. Likewise the probability of using die2 is 0.5. Now the probability of red face turns up while using die1 is $\frac46$. So the probability of red face will turn up on the third throw given the fact on the first two throws, red face turned up is A)$0.5*\frac{(\frac{4}{6})^3}{(\frac{4}{6})^3+(\frac{2}{6})*(\frac{4}{6})^2}.$
Likewise the probability of red face turns up while using die2 is B)$0.5*\frac{(\frac{2}{6})^3}{(\frac{2}{6})^3)+(\frac{2}{6})^2*(\frac46)}$
By adding both A) and B) we can get the answer to our first question.
Similarly,the probability of die1 is being used, if on the first two throws, red faces turned up is 0.27777777.[0.5*$((\frac{4}{6})^2+(\frac{2}{6})^2)$]