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Compound Probability With and Without Replacement

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Basics on the topic Compound Probability With and Without Replacement

Compound Probability with and without Replacement

Learn about how to find the probability of compound events both with and without replacement.

Transcript Compound Probability With and Without Replacement

Compound probability with and without replacement. If you recall, compound probability is the chance of two or more events happening together. An example of this would be if we had a fair coin, and a number cube and were finding the probability of flipping tails and rolling a three. We would first find the probability of each event independently. Then, multiply each value to arrive at a one out of twelve chance of the event occurring. However, sometimes with compound probability, the first event has an effect on the following event, which is called dependent probability. Let's look at this concept, with a big bowl of candy! Here, we have ten pieces of candy in total. If someone reaches into the bowl without looking, what is the probability that they will first choose a chocolate bar, replace it and then choose a lollipop? Notice the directions say that the first piece of candy is replaced, or put back before the second is chosen, so these events are independent. We can first find the probability of this compound event as two separate events. The probability of choosing a chocolate bar is two out of ten, and then choosing a lollipop is four out of ten. When we simplify, and then multiply the values, the final probability of choosing a chocolate bar, replacing it, and then choosing a lollipop is a two out of twenty-five chance. Now if someone reaches in without looking, what is the probability that they will first choose a chocolate bar, not replace it, and then choose a lollipop? This is a classic example of dependent probability. The probability of choosing the first chocolate bar is two out of ten, just like the first example. But, since this chocolate bar is not going back in, our bowl of candy is slightly different now! The probability of choosing a lollipop from the bowl of candy is now four out of nine because one piece from the total was removed. Simplify and then multiply these values to find the final probability of four out of forty-five. Back to the full bowl of candy, what is the probability of choosing a piece of bubble gum, not replacing it, and then choosing another piece of bubble gum? The probability of the first piece is three out of ten. And this time, since we are removing a piece of bubble gum on our first pick and not putting it back, we must subtract one from the total candy and the pieces of bubble gum. The probability of the second piece will be two out of nine. Multiply these together, and then simplify to see that there is a one out of fifteen chance of picking two pieces of bubble gum without replacement. For one last example, let's find the probability of choosing a lollipop, and then a chocolate bar, and then another chocolate bar without replacement. That's right, this time we have three events, without replacement. First up, the lollipop has a four out of ten chance. Then this was removed, so the probability of a chocolate bar is two out of nine. And keep that chocolate bar out, and the probability of getting another chocolate bar will be one out of eight. Simplify and multiply to arrive at a total probability of one out of ninety, which is very unlikely! Let's summarize what we have learned. If the problem says you are replacing it, then the events are independent and do not have an effect on each other. If the problem says without replacement, then it is a dependent probability, and one event will affect the others. Be sure to always read the directions carefully oh, and follow that golden rule: you touch it, you eat it!