I am working through a probability question and I'm not sure how to proceed with this problem. A bag contains $4$ marbles and at least half of the marbles are blue and the rest are red. When two balls are drawn at random without replacement, the probability of drawing two marbles of the same color is equal to the probability of drawing two marbles of different colors. How many blue marbles are in the bag?
So the possibilities for the combination of marbles in the bag are:
- $2$ Blue and $2$ Red
- $3$ Blue and $1$ Red
- $4$ Blue and $0$ Red
The possible Picks are:
- If there is $2$ Blue and $2$ Red, the possible picks will be BB, RB, or RR.
If there is $3$ Blue and $1$ Red, the possible picks will be BB or RB.
If there is $4$ Blue and $0$ Read, the possible picks will be BB.
Since we know that the probabilities are equal for picking marbles with same and different colors, the bag appears to have $3$ Blue marbles. However, I'm not sure if that is correct because I don't understand how there can be equal probability when Blue has more marbles than Red. Wouldn't it always be more likely to pick $2$ same colored (Blue) marbles since there are more in the bag?
What you did is correct, but I will explain it:
To verify that $3$ blue, $1$ red is the answer, let the blue balls be $b_1;b_2;b_3$ and the red is $r_1$.
The number of ways to choose them is $6$, because $\dfrac{4!}{2!\times 2!}=6$.
The number of ways to choose two blues from $b_1;b_2;b_3$ is $3$, half of $6$.
The number of ways to NOT choose two blues consist of the number of ways to choose two reds (which is zero) and the number of ways to choose two balls of different colors (which is $3$).
Your answer is correct and note that the problem states that there must be at least $2$ blue balls.