For a carnival game, containers are arranged in a triangular display. The top row has 1 container. The second row has 2 containe
2 answers:
You might be interested in
DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!
<h2>(a)</h2>Case 1: both balls are white.
At the beginning we have balls. We want to pick a white one, so we have a probability of
of picking a white one.
If this happens, we're left with white balls and still
black balls, for a total of
balls. So, now, the probability of picking a white ball is
The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability
Case 2: both balls are black
The exact same logic leads to a probability of
These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is
Case 1: both balls are white.
In this case, nothing changes between the two picks. So, you have a probability of of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability
of picking a black ball with both picks.
This leads to an overall probability of
Of picking two balls of the same colour.
<h2>(c)</h2>We want to prove that
Expading all squares and products, this translates to
As you can see, this inequality comes in the form
With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:
And this is our case, because in our case we have
so, y has an extra piece and it is larger
which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2