In a certain day care class, $30\%$ of the children have grey eyes, $50\%$ of them have blue and the other $20\%$'s eyes are in other colors. One day they play a game together. In the first run, $65\%$ of the grey eye ones, $82\%$ of the blue eyed ones and $50\%$ of the children with other eye color were selected. Now, if a child is selected randomly from the class, and we know that he/she was not in the first game, what is the probability that the child has blue eyes?
My solution
Let's say $B =$ blue, $G =$ grey and $O =$ "Other color" and $NR =$ "not selected for the first run"
$$P(B \mid NR) = \frac{P(NR \mid B)P(B)}{P(G)P(NR \mid G) + P(B)P(NR \mid B) + P(O)P(NR \mid O)}$$
On substituting values $$P(B \mid NR) = \frac{0.5 \cdot (1-0.82)}{(0.3 \cdot (1-0.65)) + (0.5 \cdot (1-0.82)) + (0.2 \cdot (1-0.5))}$$
$$P(B \mid NR) = 0.305$$
Is this the right way to use bayes theorem?
