Comparing results after performing W/ and W/O replacement on an experiment In this is famous example in the probability theory, there are 6-Red, 4-Green, and 5-Blue balls in a bag. By calculating the probability with and without replacement for these three colors, we multiply as:


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*With Replacement: $6/15 * 4/15 * 5/15 = 0.036$

*Without Replacement: $6/15 * 4/14 * 5/13 = 0.044$
My question is: Why does Without Replacement tend to produce larger value than With Replacement even though the balls are taken out of the bag in every step of the experiment?
 A: This statement is not always true.  For instance, if we are looking to select 3 green balls, then


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*With replacement it is: (4/15)^3 = 64/15^3

*Without replacement it is: 4/15 * 3/14 * 2/13 = 24 / 15*14*13, clearly smaller
The reason in your case it produces a larger value is because, taking out a marble of one color makes the probability of selecting the next marble of a different color higher.
So if you are looking for selecting marbles of the same color, the With replacement will always be higher; if you are always selecting marbles of a different color, the Without replacement will always be higher.
A: I guessing your question is the probability of drawing red green blue in each scenario. The differences is due to the decreasing denominator.
In the first case, the probability is
$$\frac{6}{15}\cdot\frac{4}{15}\cdot\frac{5}{15} = \frac{8}{225}.$$
In the second scenario
$$\frac{6}{15}\cdot\frac{4}{14}\cdot\frac{5}{13} = \frac{4}{91}.$$
Getting a common denominator gives
$$\frac{728}{20475} < \frac{900}{20475}.$$
Take for example a different scenario. You have a box full of 8 balls, 4 red and 4 green. With replacement, you replace the ball after a draw. So the probability of drawing a red, then a green is
$$\frac{4}{8}\cdot \frac{4}{8} = \frac{16}{64} = \frac{112}{448}.$$
Without replacement, you do not replace the ball. Thus,
the probability of red then green is
$$\frac{4}{8}\cdot \frac{4}{7} = \frac{2}{7}= \frac{124}{448}.$$
since there were 8 balls at first, then there were 7 balls in the second draw.
Notice that in the second case, it was easier to draw a green because there were fewer balls total in the box.
Since it was easier, the probability is higher and thus
$$\frac{112}{448}<\frac{124}{448}.$$
This same logic applies to the original scenario.
A: It produces a bigger number BECAUSE the balls are being taken out. Think of it this way:
When you take out a red ball, a larger portion of the remaining balls are green. this is visible in your equations because $\frac{4}{14}>\frac{4}{15}$. dividing by a smaller number gives a bigger number.
