Poor Andrew in a Mental Hospital
In a certain mental hospital, poor Andrew works as a janitor. He observes that 30 % of the patients have grey eyes, 50 % have blue eyes and the remaining 20 % have eyes that are of other colors. One day they play a game together to irritate Andrew. In the first run, 65 % of the grey eyed ones, 82 % of the blue eyed ones and 50 % of other eye colored ones get together. Now, if a patient is selected randomly from the mental hospital, and we know that he/she was not in the first run of the game, what is the probability that the patient has blue eyes?
Answer required to 3 decimal places.
The answer is 0.305.
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1 solution
Let there be n patients.
Grey eyed patients not in the first run = n ⋅ 0 . 3 ⋅ ( 1 − 0 . 6 5 ) = 0 . 1 0 5 n
Blue eyed patients not in the first run = n ⋅ 0 . 5 ⋅ ( 1 − 0 . 8 2 ) = 0 . 0 9 n
Other eyed patients not in the first run = n ⋅ 0 . 2 ⋅ ( 1 − 0 . 5 ) = 0 . 1 n
Total number of patients not in the first run = 0 . 1 0 5 n + 0 . 0 9 n + 0 . 1 n = 0 . 2 9 5 n
Therefore, P ( blue eyes given not in first run ) = 0 . 2 9 5 n 0 . 0 9 n = 0 . 3 0 5 (to 3.d.p).