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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?

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up vote 3 down vote accepted

Yes, this is correct.

The general form of Bayes' rule is

$$P(A_i|B)= \frac{P(B|A_i)P(A_i)}{\sum_jP(B|A_j)P(A_j)}$$

as you've used above.

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Thanks for the generalized version of Bayes' theorem. Do you see anything wrong with the way I have arrived at the individual probabilities? – naresh Nov 16 '12 at 17:12
Nothing wrong that I can see... – Simon Hayward Nov 16 '12 at 18:49
Has that helped, do you have the answer now? – Simon Hayward Nov 16 '12 at 19:23

$$p(\textrm{not game 1} \mid \textrm{blue}) \div p(\textrm{not game 1}) = 0.18×0.50 \div (0.18×0.50+0.5×0.2+0.35×0.30) =0.3051$$

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