Consider 2 identical boxes $B_1,B_2$, where $B_i,i=1,2$ contains $i+2$ red and $5-i-1$ white balls. A fair die is cast. Let the number of dots shown on the top face of die be $N$. If $N$ is even or 5, then 2 balls are drawn with replacement from $B_1$, otherwise, 2 balls are drawn with replacement from $B_2$. The probability that the 2 balls drawn are of different colors is $1. \frac{7}{25} 2. \frac{9}{25} 3. \frac{12}{25} 4. \frac{16}{25}$
Given answer: 3
My attempt:
$B_1$ contains 3 red and 3 white balls, 6 total. $B_2$ contains 4 red and 2 white balls, 6 total.
$B_1$ is chosen if $N=2,4,6,5$, so $P(B_1)=\frac46=\frac23$ and $P(B_2)=\frac26=\frac13$
Let $E$ be the event that the 2 balls are of different colors.
Then $P(E)=P(E \cap B_1) + P(E \cap B_2)$
$=P(B_1) P(E|B_1) + P(B_2) P(E|B_2)$
$=\frac23 \left[\frac{3}{6} \frac{3}{6} + \frac{3}{6} \frac{3}{6}\right]+\frac13 \left[\frac{4}{6} \frac{2}{6} + \frac{2}{6} \frac{4}{6}\right]=0.62$
This is using the logic that one can pick $3$ out of $6$ red balls and $3$ out of $6$ white balls from $B_1$ with replacement, and this can happen in the order red, white or white, red.
Where did I go wrong?