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# Probability Without Replacement – Explanation & Examples

Did you ever take candy from a box, gave it to your older kid (who ate the candy), then took a second candy, and gave it to your younger kid? If your answer is yes, you have already come across a nice story involving dependent probability, generally termed as **Probability Without Replacement. **But if your older kid had returned the candy and you had put it back in the box, the **second event** would have been **independent**. The box of sweets (sample space) would have remained the same for the second event — **probability with replacement.**

*Probability without replacement involves dependent events where the preceding event has an effect on the probability of the next event.*

**Probability Without Replacement **may initially sound tricky but trust me; it is one of the most specific mathematics topics. This lesson will clear your concept about dependent probability, and we will learn step by step how to calculate probability without replacement. This lesson will also help you understand how to use a probability tree diagram involving dependent probability — probability without replacement.

**What does probability without replacement mean?**

**Probability without replacement** means once we draw an item, then we **do not replace** it back to the sample space before drawing a second item. In other words, an item **cannot be drawn more than once.**

For example, if we draw a candy from a box of 9 candies, and then we draw a second candy without replacing the first candy. Of course, the sample space would no longer remain 9 for the second event because we have not replaced the first candy. Thus, the sample space would be 8 for the second event. In other words, the sample space has been **changed** for the second event.

We use **probability without replacement** to solve the problems where the **sample space changes** for different events and the occurrence of the next event **depends upon** what happens in the preceding event.

**How to calculate probability without replacement?**

Now that we briefly understand dependent probability – before diving deep into calculating probability without replacement – let’s first try to visualize what **dependent events** are and how the preceding event can affect the next event.

For example, a box contains $4$ green candies and $5$ brown candies, as shown below.

Let us consider that event A is **getting a green candy**. As there are $4$ green candies, so the **number of ways it can happen = **$4$. Also, the total number of candies is $9$, so the **total number of outcomes = **$9$.

Let’s suppose $P(A)$ represents the **probability of** **getting a green candy**. Thus, we can calculate the probability of getting a green candy such as:

**P(A) = number of ways it can happen **$\displaystyle /$** total number of outcomes**

$P(A) = \frac{4}{9}$

Thus, the probability of getting a green candy is $4$** in **$9$**, **as shown below.

Now going ahead, we need to understand that the probabilities **change** if we take any candy out.

Here, let us check the two scenarios such as:

**Scenario 1:**

If we draw a **brown** candy before, then the probability of a green candy next is $4$** in **$8$**, **as shown below.

**Scenario 2:**

And if we draw a **green** candy before, then the probability of a green candy next is $3$** in **$8$**, **as shown below.

Here is the complete snapshot of what we learned.

The probabilities **change** because we are **removing** candies from the box. The next event tends to depend on what occurs in the preceding event, also known as a **dependent.**

Thus, when we remove the candies each time **without replacement**, then the probabilities **change,** and the events are **dependent.**

A **tree diagram** is a pleasing way to visualize the concept involving probability without replacement.

Taking the same example, there is a $\frac{5}{9}$ chance of drawing a brown candy from the box, and a $\frac{4}{9}$ chance for green candy such as:

Going ahead, let’s check what occurs when we draw a second candy.

If a **brown** candy is chosen first, there is a $\frac{4}{8}$ probability of getting a brown candy and a $\frac{4}{8}$ probability of getting a green candy. The reason is that after $1$ brown candy is taken, we have $8$ candies left, of which $4$ are brown, and $4$ are green.

If a **green** candy is chosen first, there is a $\frac{5}{8}$ probability of getting a brown candy and a $\frac{3}{8}$ probability of getting a green candy. Again, after $1$ green candy is taken, we have $8$ candies left, of which $3$ are brown, and $5$ are green.

Now, we can determine a variety of **probabilities without replacement** with the help of the above tree diagram.

**For example,** let’s say we need to determine the probability of drawing $2$ green candies. We can visualize from the tree diagram that it is a $\frac{4}{9}$** probability** followed by a $\frac{3}{8}$** probability, **as shown in the tree diagram below.

From the tree diagram, we can easily figure out that we have just multiplied the probabilities such as:

$\frac{4}{9}\times \frac{3}{8}=\frac{12}{72}=\frac{1}{6}$

Thus, the **probability of drawing 2 green candies** is $\frac{1}{6}$.

**Probability without replacement formula**

In our example, event $A$ is **getting a green candy,** and $P(A)$ represents the **probability of getting a green candy** with a probability of $\frac{4}{9}$:

$P(A) = \frac{4}{9}$

Also, event $B$ is **getting a green candy second, **but for that, we have two scenarios such as:

- If we chose a
**green**candy first, the probability is now $\frac{3}{8}$. - If we chose a
**brown**candy first, the probability is now $\frac{4}{8}$.

It is up to us which one we chose. And we can use the notation $P(B\mid A)$ read as** ‘the probability of **$B$** given **$A$**‘. **Please note that the symbol ‘$|