Conditional Probability
I have a probability question, and a solution for it. I also have an incorrect solution, and I want to be sure of the reason that it is incorrect. The question is as follows:
A box contains 10 red and 5 green marbles. One six-sided die is rolled. If the outcome of the roll is an even number, that many red marbles are added to the box. If the outcome of the roll is an odd number, that many green marbles are added to the box. Finally, one marble is selected from the box. What is the probability that the marble selected is red?
If I roll:
1 -> 10 red, 6 green -> 10/16 probability
2 -> 12 red, 5 green -> 12/17 probability
3 -> 10 red, 8 green -> 10/18 probability
4 -> 14 red, 5 green -> 14/19 probability
5 -> 10 red, 10 green -> 1/2 probability
6 -> 16 red, 5 green -> 16/21 probability
The probability I get to each of those outcomes is 1/6, so the probability of getting a red marble is (1/6)*(10/16 + 12/17 + 10/18 + 14/19 + 1/2 + 16/21) = 0.648. I believe this to be the correct solution.
However, I noticed that there are 111 marbles in total and 72 of them are red. 72/111 = 0.649. Is it just coincidence that the numbers are so close? I know that you can only use the formula P(A) = (number of outcomes that make up A / all possible outcomes) if the outcomes are equally likely. Here, I have 6 different possibilities and in each of those, the probability of drawing a red marble is not equally likely. I'd love some insight into this problem.
A box contains 10 red and 5 green marbles. One six-sided die is rolled. If the outcome of the roll is an even number, that many red marbles are added to the box. If the outcome of the roll is an odd number, that many green marbles are added to the box. Finally, one marble is selected from the box. What is the probability that the marble selected is red?
If I roll:
1 -> 10 red, 6 green -> 10/16 probability
2 -> 12 red, 5 green -> 12/17 probability
3 -> 10 red, 8 green -> 10/18 probability
4 -> 14 red, 5 green -> 14/19 probability
5 -> 10 red, 10 green -> 1/2 probability
6 -> 16 red, 5 green -> 16/21 probability
The probability I get to each of those outcomes is 1/6, so the probability of getting a red marble is (1/6)*(10/16 + 12/17 + 10/18 + 14/19 + 1/2 + 16/21) = 0.648. I believe this to be the correct solution.
However, I noticed that there are 111 marbles in total and 72 of them are red. 72/111 = 0.649. Is it just coincidence that the numbers are so close? I know that you can only use the formula P(A) = (number of outcomes that make up A / all possible outcomes) if the outcomes are equally likely. Here, I have 6 different possibilities and in each of those, the probability of drawing a red marble is not equally likely. I'd love some insight into this problem.