# If 3 jelly beans are taken from the above question without replacement, what is the probability that the first is red, the second is black and the third is yellow?

The question is incomplete. So I have made up some numbers.

This is more a case of demonstrating the method.

The final probability is the product of the 3

In a lot of cases probability is a matter counting. Some scenarios are a lot more complex than others.

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The probability that the first jelly bean drawn is red, the second is black, and the third is yellow, without replacement, can be calculated as the product of the probabilities of each event.

Let's assume there are a total of 5 jelly beans: 1 red, 1 black, 1 yellow, and 2 other colors.

The probability of drawing a red jelly bean first is 1/5. After removing a red jelly bean, there are 4 jelly beans left, so the probability of drawing a black jelly bean second is 1/4. After removing a black jelly bean, there are 3 jelly beans left, so the probability of drawing a yellow jelly bean third is 1/3.

Multiplying these probabilities together: 1/5 * 1/4 * 1/3 = 1/60

So, the probability that the first jelly bean is red, the second is black, and the third is yellow is 1/60.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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