Probability of selecting a combination of two variables.

Question

I have a bag of toys. 10% of the toys are balls. 10% of the toys are blue.

If I draw one toy at random, what're the odds I'll draw a blue ball?

Answer 1

Since it is not entirely clear from the question, I will assume every toy has an equal probability of being blue. So we are assuming the unlikely proposition that no correlation (positive or otherwise) exists between color and type-of-toy. Otherwise, could very well be that 10% of the toys are neon-green balls (tennis balls, e.g.) and 10% of the toys are blue blocks, in which case you have 0% probability that you'll draw a blue ball.

We know $10$% $ = 0.1$ of the toys are balls, and $10$% $= 0.1$ of these balls are blue.

Then $10$% of ($10$%) of the toys are blue balls.

So the probability of drawing a blue ball is $0.1 \times 0.1 = 0.01$.

This equates to a $0.01 \times 100$% = $1$% probability of drawing a blue ball.

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Answer 2

Assuming that the balls are not any more or less likely to be blue that other toys, the object drawn has a $.1$ chance of being a ball and a $.1$ chance of being blue. So the chance of it being a blue ball is $.01$.

Answer 3

If the "blueness" is uniformly distributed. We have S Toys

From wich $0.1 \mathbf S$ are Balls.
$P(X)= \frac {0.1S}S=0.1$

$0,1 \mathbf S$ are Blue toys.
$P(Y)=\frac {0.1S}S=0.1$

If you want Blue ball you have
$P(XY)=P(X)\cdot P(Y) = 0.1 \times 0.1 = 0.01 = 1 $%