# What is the probability of selecting at least one each?

There are three marbles in the bag: Green, Yellow, Blue.

Probability of,
Green: $\frac{1}{4}$
Yellow: $\frac{1}{2}$
Blue: $\frac{1}{4}$

Four marbles are drawn with replacement, what is the probability of getting at least one of each?

My work:
$$\frac{1}{4}*\frac{1}{2}*\frac{1}{4}*\frac{1}{4}$$ $$+\frac{1}{4}*\frac{1}{2}*\frac{1}{4}*\frac{1}{2}$$$$+\frac{1}{4}*\frac{1}{2}*\frac{1}{4}*\frac{1}{4}$$

This problem seems so easy, but I just can't find a book with good reference to confirm my work. I hope this is correct.

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"getting at least one of each" when you do what? Is it that you pick a marble three times, replacing it each time? The problem needs some specification of what you want the probability of. –  coffeemath Nov 8 '13 at 1:35
Yes, what you said is correct. I have added the specification. –  user13985 Nov 8 '13 at 1:44
You should also include the specification of drawing three times in the question itself. Not everyone will read all the comments. Maybe rephrase as "What is the probability of getting each marble once in three selections with replacement?" [Note that "at least once" is the same as "once" in this case of three marbles and three draws.] –  coffeemath Nov 8 '13 at 1:53
I changed it now. –  user13985 Nov 8 '13 at 2:09

Another way of solving the problem is to get the following combination

For the probabilities to make sense, the minimum marbles that got there are two of a kind and one of a kind of the other two.

Given that the combinations that you will draw them will be the following four draws (2G,1Y,1B), (1G, 2Y, 1B) and (1G, 1Y, 2B).

For each one of them,

P(2G,1Y,1B) = (1/4)^2*(1/2)(1/4)(4!/2!) - Meaning 2G,1Y,and 1B could have 12 different

ways to appear in the four draws.

Similarly, P(1G, 2Y, 1B) = ((1/4)(1/2)^2(1/4)*(4!/2!) - The reasoning is the same.

Similarly, P(1G, 1Y, 2B) = (1/4)(1/2)(1/4)^2*(4!/2!) - The reasoning is the same

Adding them up all is not that different from your [(1/128)+(1/64)+(1/128)]*(4!/2!)

= [1/32]*12 = 3/8

Thanks

Satish

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As you rightly point out, the OP answer would be right if it were multiplied by $4!/2!$, which is the combinations of four things if two of them are the same. –  coffeemath Nov 8 '13 at 12:38
No Problem, Thanks, Satish –  satish ramanathan Nov 8 '13 at 13:31
The bag may be viewed as $\{G,Y_1,Y_2,B\}$ where the subscripts on the $Y$ are to temporarily distinguish the two yellow marbles. This gives your probabilities of $1/4,1/2,1/4$ for $G,Y,B$ (green, yellow, blue) on each draw.
Then four draws can result in any of $4*4*4*4=4^4$ outcomes, and each is equally likely. Now how many combinations contain at least one of each color marble? [With one previous version suggested in a comment, the number of draws was three, and this part was easier.]
We can either have 2 green and one each of yellow, blue, or one green, two yellow, and one blue, or finally one green one yellow, and two blue. For each of these one has to find the number of ways, then total all, and place over $4^4$ for the final probability.
For example two green and one each of yellow, blue is either of $G,G,Y_1,B$ or $G,G,Y_2,B$ and both of these can occur in 12 ways (since order matters in the sample space). I'll leave the other cases for you to look at. But note I don't think this problem is as simple as your attempt.
ADDED: I finished the calculations above and got $3/8$ as the final probability of getting at least one of each colors in four drawings with replacement. This differs from the value computed in the posted question. The sum of those three products of fractions is $1/32.$