# Probability of choose the same marble again

Box I contains 4 red and 8 blue marbles while box II contains 5 red and 3 blue marbles. An unfair coin is tossed – whose probability of turning up heads is 40%. If the coin comes up heads box I is chosen and a random marble is chosen, otherwise if it is tails the marble is chosen from box II.

Suppose after the first marble is chosen – the experiment is repeated. Assume the first marble is NOT put back into its box. The coin is flipped again and another marble is chosen from either box I or box II.

(d) What is the probability that the second marble has the same color as the first marble?

I feel like the answer is so obvious, but I can't get it. In the first case $P(Red)$ = .5083 while $P(Blue)$ = 4917. I'm just confused after this point whether where I go after this.

• The probability of $[RI ,RI ]$ is $0.4 \cdot\frac{4}{4+8}\cdot 0.4 \cdot\frac{4-1}{4-1+8}=\frac{ 4}{275}$
• The probability of $[RI ,RII]$ is $0.4 \cdot\frac{4}{4+8}\cdot(1-0.4)\cdot\frac{5 }{5 +3}=\frac{ 1}{ 20}$
• The probability of $[RII,RI ]$ is $(1-0.4)\cdot\frac{5}{5+3}\cdot 0.4 \cdot\frac{4 }{4 +8}=\frac{ 1}{ 20}$
• The probability of $[RII,RII]$ is $(1-0.4)\cdot\frac{5}{5+3}\cdot(1-0.4)\cdot\frac{5-1}{5-1+3}=\frac{ 9}{ 70}$
• The probability of $[BI ,BI ]$ is $0.4 \cdot\frac{8}{4+8}\cdot 0.4 \cdot\frac{8-1}{4+8-1}=\frac{56}{825}$
• The probability of $[BI ,BII]$ is $0.4 \cdot\frac{8}{4+8}\cdot(1-0.4)\cdot\frac{3 }{5 +3}=\frac{ 3}{ 50}$
• The probability of $[BII,BI ]$ is $(1-0.4)\cdot\frac{3}{5+3}\cdot 0.4 \cdot\frac{8 }{4 +8}=\frac{ 3}{ 50}$
• The probability of $[BII,BII]$ is $(1-0.4)\cdot\frac{3}{5+3}\cdot(1-0.4)\cdot\frac{3-1}{5+3-1}=\frac{27}{700}$