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This question is a probability question from a practice paper of a Statistics module that I have to take as part of my university graduation requirements.

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In a certain day care class, 30% of the children have brown eyes, 20% of them have blue eyes and the other 50% have eyes that are in other colours. One day, some of them play a game together. In the game, 45% of the children have brown eyes, 20% have blue eyes and 35% have other eye colours. Now, if a child is selected randomly from the class, and we know that he/she was not in the game, what is the probability that the child has blue eyes (rounding off to 2 decimal places)?

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My attempt at answering this question:

Let the total number of children in the class be x.

Let the total number of children who played the game together be y.

Total brown eyes. blue eyes and other coloured eyes respectively: 0.3x, 0.2x, 0.5x

Total children (in game) with brown eyes, blue eyes and coloured eyes respectively: 0.45y, 0.2y, 0.35y

P(Blue Eyes | Not In Game) = P(Blue Eyes) * P(Not In Game) = 0.2x/(0.2x+0.3x+0.5x) * (x-y)

How do I get an absolute value as an answer since there are so many unknowns? (Biggest one of them being, we don't know exactly how many students are in the game)

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You do not know how many children there are in total (your $x$) and how many play the game (your $y$). But you may be able to answer the specific question anyway

Let's set up a table of children playing and eye colour:

         All      Playing     Not playing
Brown   0.30 x    0.45 y     0.30 x - 0.45 y 
Blue    0.20 x    0.20 y     0.20 x - 0.20 y  
Other   0.50 x    0.35 y     0.50 x - 0.35 y
        ======    ======     ===============
Total     x         y             x - y

So if a child is selected randomly from the class not playing the game, the probability that the child has blue eyes is $\dfrac{0.20x - 0.20 y}{x-y}=0.20$

This would not work with Brown eyes as you could not simplify $\frac{0.30x - 0.45 y}{x-y}$

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