# Can You Solve This Intro Probability Problem?

An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is 0.44.

Calculate the number of blue balls in the second urn.

– This problem is from beanactuary.org.

# Basic Probability Rules

If this is your first run at probability, *kudos to you! *Probabilities aren’t only a fundamental component of Statistics but also an inherent aspect of everyday life. This problem will utilize a few fundamental rules of probability, but first let’s define probability.

## Definition

*Probability** is a measure of the likelihood of a desired event(s) happening.*

*We obtain this value by dividing the number of possible ways of achieving the desired event by the total number of possible events.*

Since the total number of events is always equal to or larger than the desired event, **probabilities are always a number between 0 and 1. **If you obtain a number not within that range, check your arithmetic!

## Example

**Suppose we want to find the likelihood of drawing a blue ball from the first urn described above. **Simply divide the number of blue balls in the first urn by the total number of balls in the urn.

The chance of drawing a blue ball from the first urn is 6/10 or 0.60.

# Combining Probabilities

This is where it gets a little tricky. When we have multiple event probabilities, as in the problem above, we have to think about *how* to combine them. In general, there are two scenarios: *the AND scenario, the OR scenario.*

## the AND scenario

In this scenario, we want more than one event to happen together. For example, drawing two red balls from the two urns described above is the same as saying,

*“I want to a red ball from the first urn **AND** a red ball from the second urn.”*

Here I have two distinct probabilities to calculate:

- Drawing a red ball from the first urn.
- Drawing a red ball from the second urn.

Once I calculate these probabilities,** I multiply them together** to obtain the combined “AND” probability.

## the OR scenario

In the OR scenario, we recognize events or cases that could happen, but don’t all have to happen together. For example, in the above problem it tells us the probability of the both balls being the same color is 0.44. This can be thought of as:

*“Drawing two red balls **OR** two blue balls results in a probability of 0.44.”*

The “OR” separates our calculation into two parts:

**Drawing two red balls**(i.e drawing a red from the first urn AND a red from the second.)**Drawing two blue balls**(i.e. drawing a blue from the first AND a blue from the second.

Once I have calculated the two separate probabilities, **I add them together **to find the combined “OR” probability.

# Solution

*“The probability that both balls are the same color is 0.44,” *is the same as saying, “*Drawing two red balls **OR** two blue balls results in a probability of 0.44.” *Replace *drawing two red balls and drawing two blue balls* with the AND statements we learned above:

(Drawing a red ball from urn 1ANDa red from urn 2)

OR

(drawing a blue ball from urn 1ANDa blue from urn 2).

*Drawing a red ball from urn 1 AND a red ball from urn 2.*

*Drawing a red ball from urn 1 AND a red ball from urn 2.*

To solve this probability determine the individual probabilities of drawing a red ball from each respective urn.

The probability of drawing a red ball from urn 1 is 4/10.

Calculating the probability of red balls from urn 2 is a tad trickier. We know there are 16 red balls in the second urn, but we don’t know the total number of balls in the urn because we’re unsure of how many blue balls it holds.

*That’s okay!* The point of the problem is to figure out the number of blue balls in urn 2, so for the meantime we’ll *let a variable represent the blue balls.*

Now write the expression for the probability of drawing a red ball from the second urn. (Note: the total number of balls in the urn is the RED + BLUE or in this case 16 + x.)

Since this is an AND scenario, multiply the two fractions together.

## Drawing a blue ball from urn 1 AND a blue ball from urn 2.

We’ll leave that expression alone for a moment, and skip to the second AND scenario. Like above we calculate the two individual probabilities using *x* for the number of blue balls in urn 2.

This is another AND scenario, so multiply the two fractions together.

## Implementing the OR scenario

Recall our original objective is writing an expression for this statement:

*(**Drawing a red ball from urn 1** **AND a red from urn 2**) OR (**drawing a blue ball from urn 1 AND a blue from urn 2**) = 0.44**.*

Now that we have the two AND scenarios, we can add them together, set equal to 0.44 and solve for *x, the number of blue balls in urn 2*.

*Yay, we finally have our expression to solve!* Start by multiplying the fractions together (I’ll leave the denominators undistributed for the time being).

Since we have a common denominator, add the fractions together.

Now multiply by 10(16 + x) on both sides to eliminate the fraction.

Which leaves us with:

Finish the algebra to solve for x.

Therefore, the second urn must contain** 4 blue balls.**

*Next → **The “Zero Power Rule” Explained*

*Thanks for reading!*

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