Виборки з поверненням

$$\textbf{Probability: Selecting with Replacement}$$

When you always replace or put back selected items, calculating probability can be pretty straightforward!

Notice how when you replace a marble, the amount of marbles in the bag stays the same, and thus the denominator of your fraction stays at $7.$

Let's do another example with marbles. Here are refreshers on theoretical probability and probability of independent events to help you if you need it.

Since there are $4$ pink marbles in the bag and $7$ marbles total, the probability of choosing a pink marble is $\dfrac{4}{7}.$ In other words: $$P(\text{pink}) = \dfrac{4}{7}$$

Then, you replace whatever marble you drew, so there are still $7$ marbles in the bag.

Next, since there are $2$ orange marbles in the bag and $7$ marbles total, the probability of choosing an orange marble is $\dfrac{2}{7}.$ In other words: $$P(\text{orange}) = \dfrac{2}{7}$$

Then, you replace whatever marble you drew, so there are still $7$ marbles in the bag.

Finally, since there are still $4$ pink marbles in the bag and $7$ marbles total, the probability of choosing a pink marble is $\dfrac{4}{7}.$ In other words: $$P(\text{pink}) = \dfrac{4}{7}$$

Remember, all these events are independent since you always replaced the marble you drew, and the probability of picking any one marble will stay the same between selections.

Thus, to find the probability of drawing a pink marble, replacing it, drawing an orange marble, replacing it, and finally drawing another pink marble is equal to the probability of drawing a pink marble times the probability of drawing an orange marble times the probability of drawing a pink marble. So: $$\begin{align} P(\text{pink, then orange, then pink}) &= \dfrac{4}{7} \cdot \dfrac{2}{7} \cdot \dfrac{4}{7}\\ &=\dfrac{4\cdot 2 \cdot 4}{7 \cdot 7 \cdot 7}\\ &=\dfrac{32}{343} \end{align}$$

Selection with Replacement

Selecting with replacement is exactly what it sounds like - you're picking something out of a bag (bucket, drawer, group, etc.), putting it back in there (i.e., the replacement), and picking another one of those somethings out.

Then, after all that selecting and replacing, you calculate probabilities. Sounds riveting, right?! Sarcasm aside, replacement problems are pretty easy to navigate and solve.

Notice that the denominator of our fraction remains the same. This is because we have the same total number of paperclips in the box, thanks to replacement.

If I have a bag of marbles, $10$ red and $20$ white, and I pick one out, record the color and put it back in before I pick my second, that is called selecting with replacement. However, if I first pick out one marble, set it aside, and later pick out another marble from the $29$ - marble bag, that is called selecting without replacement.

When you see a probability problem, always see if the question calls for replacement.

(Video) Probability with Replacement

by Kevin Dorey

Kevin Dorey has a great video explaining probability with replacement.