A bag contains equal number of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it, what is the probability that at least one will be red?

I know that we can solve this very easily using the probability of this not happening, for eg. there is a 2/3 probability that the marble will not be red. so the probability of all three marbles not being red is 2/3 * 2/3 * 2/3 = 8/27.

so the probability of atleast one marble is 1 - 8/27 = 19/27.

But I want to solve this problem by actually calculating the probability of this happening one by one, how do I do it?

I tried doing this like this but the answer came out incorrect: 1/3 + 1/3 * 1/3 + 1/3 * 1/3 * 1/3 = 13/27. which is incorrect.

Requesting any help with this, Thank you.


To do this the additive way, you'll need to calculate the probability of getting exactly one red marble, exactly two red marbles, and exactly three red marbles. These you can add up because they're disjoint sets of outcomes.

If the first marble is red, then the second and third aren't: $1/3 \times 2/3 \times 2/3 = 4/27$. Same for only the second marble red, and for only the third marble red, for a total of $12/27$.

Can you finish from here?


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