# Probability Questions and Answers: Brush Up on Your Probability Skills

In probability, you explore the likelihood of something occurring. A very common probability problem is rolling a die 25 times and figuring out how many times the die hits six. You then find the probability of such an event occurring. This kind of math is not the exact, black and white questions and answers of other mathematical areas. This type involves gray area where something might happen a certain number of times, but it could also happen more or less times than stated. If you’re looking to brush up on your skills, try these sample questions and answers for probability. Refresh yourself on the formulas, and perhaps rebuild your foundation with a beginner’s probability class.

## The Probability of a Single Event

This particular formulas is used for simple probability questions like the question of rolling the six-sided die and finding out which number you would roll. In this way, we use the number of favorable outcomes over the number of possible, equally-likely outcomes to get the probability. To create this formula, we will use f for the favorable outcomes and the letter l for equally-likely outcomes. P will be used for probability. The formula would look like this:

P = f/l

Try a couple example problems to learn how to use the formula.

**Example One**

You are going to roll a six-sided die. There are six possible outcomes, and each is equally likely to occur. You want to know the probability of rolling a 3 or a 5. So, let’s use our formula from above and the numbers that we know.

P = f/l

favorable outcomes (f) = 2

equally-likely outcomes (l) = 6

P = 2/6 = 1/3

The probability of rolling a 3 or a 5 on a six-sided die is 1/3. Another way to say it is that there’s a one in three chance of rolling a 3 or 5.

**Example Two**

You are going to roll a six-sided die. There are six possible outcomes, and each is equally likely to occur. You want to know the probability of rolling a 1, a 4, or a 5. Let’s use the same formula form above and the numbers that we know.

P = f/l

favorable outcomes (f) = 3

equally-likely outcomes (l) = 6

P = 3/6 = 1/2

As you can see, the probability of getting one of your favorable outcomes increases the more outcomes you have.

## The Probability of Two or More Independent Events

Whenever you have two events you want to find the probability of, your first step should be to figure out if they’re independent events. If the probability of the second event will still occur whether or not the first event occurs, they are considered independent events. A good example of independent events is the probability of heads or tails on the second toss, no matter the input of the first coin toss. There are two formulas for independent events. Try a workshop in probability and statistics.

The probability of both Event A and Event B occurring is the product of the probabilities of each separate event. The equation is below.

P(A and B) = P(A) x P(B)

Let’s do a couple of example problems to practice this new formula.

**Example Three**

You’re going to be tossing a coin twice. There are two possible outcomes, and each is equally-likely to occur. You want to know the probability of a heads on the first toss and a tails on the second toss. Let’s use the formula from above and the numbers that we know.

P(A and B) = P(A) x P(B)

P(A) = f_{A}/l_{A}

P(B) = f_{B}/l_{B}

favorable outcomes for A (f_{A})= 1

favorable outcomes for B (f_{B})= 1

likely outcomes for A (l_{A})= 2

likely outcomes for B (l_{B})= 2

P(A) = 1/2

P(B) = 1/2

P(A and B) = 1/2 x 1/2

P(A and B) = 1/4

**Example Four**

You’re going to be tossing a coin once, and you’re going to roll a six-sided die once. The coin toss has two possible, equally-likely outcomes, and the die roll has six possible, equally-likely outcomes. You want to know the probability of a tails on the coin toss and a 5 on the die roll. Use the formula from above and the numbers that we know.

P(A and B) = P(A) x P(B)

P(A) = f_{A}/l_{A}

P(B) = f_{B}/l_{B}

f_{A} = 1

f_{B} = 1

l_{A} = 2

l_{B} = 6

P(A) = 1/2

P(B) = 1/6

P(A and B) = 1/2 x 1/6

P(A and B) = 1/12

If A and B are independent events and either A or B occurs, there are three possibilities of what occurred. It is possible that A occurred and B did not, it’s possible that A did not occur and B did, and it’s also possible that both A and B occurred. The formula is below:

P (A or B) = P(A) + P(B) – P(A and B)

Here’s a few example problems for this formula. Take an online class in probability.

**Example Five**

You’re going to flip a coin two times. You want to know the probability of getting a heads on the first or second flip or even on both. Let’s use the formula from above to figure out the problem.

P(A or B) = P(A) + P(B) – P(A and B)

P(A and B) = P(A) x P(B)

P = f/l

P(A) = 1/2

P(B) = 1/2

P(A and B) = 1/2 x 1/2 = 1/4

P(A or B) = 1/2 + 1/2 – 1/4

P(A or B) = 2/4 + 2/4 – 1/4

P(A or B) = 4/4 – 1/4

P(A or B) = 3/4

**Example Six**

You’re going to flip a coin and throw a six-sided die. You want to know the probability of getting a head on the coin flip or a six on the die or both. Again, use the formula from above.

P(A or B) = P(A) + P(B) – P(A and B)

P(A and B) = P(A) x P(B)

P = f/l

P(A) = 1/2

P(B) = 1/6

P(A and B) = 1/2 x 1/6 = 1/12

P(A or B) = 1/2 + 1/6 – 1/12

P(A or B) = 6/12 + 2/12 – 1/12

P(A or B) = 8/12 – 1/12

P(A or B) = 7/12

## Conditional Probabilities for Dependent Events

There are times when an event is dependent on the occurrence of another event. Problems like this have conditions, and those conditions have to be met in order for the problem to be solved correctly. Stating the conditions is written below:

P(B | A)

The vertical bar between the two events is read as meaning “given.” In other words, this equation means, “the probability of B given A.” This usually means that numbers used for A will be changed to fit for B. To find out the probability of such a question, you would use the formula P(A and B), but it would not be the same formula we used for independent events. Here’s the formula for dependent events:

P(A and B) = P(A) x P(B | A)

As you can see, the probability of A is not dependent on the outcome of the probability of B, but B is dependent on the outcome of A. Picking cards out of a deck is a good example of conditional probability. Let’s try some example problems.

**Example Seven**

You want to know the probability of drawing two Aces out of a deck at random. There are 52 cards in a deck, and there are 4 Aces. Find the probability using the formula above.

P(A and B) = P(A) x P(B | A)

P(A) = 4/52

P(B | A) = 3/51

*Note that the numbers for finding the probability of B have changed. This is because when A occurs, there will only be 3 Aces left in the deck, and there will only be 51 cards.*

P(A and B) = 4/52 x 3/51

P(A and B) = 12/2652 = 1/221

As you can see, the probability of drawing an Ace after another Ace is not very likely at all. Let’s try another example.

**Example Eight**

You want to know the probability of drawing a Queen of Hearts and then a black card. There are 52 cards in a deck, only one Queen of Hearts, and 26 black cards. Find the probability using the above formula.

P(A and B) = P(A) x P(B | A)

P(A) = 1/52

P(B | A) = 26/51

P(A and B) = 1/52 x 26/51

P(A and B) = 26/2652 = 1/102

### Important Things to Remember About Conditional Probability Problems

Whenever working out the probability of event B given event A occurs first, you’re going to have to chance certain numbers depending on what event A is. If you are picking marbles out of a bag and want to know the probability of picking out a red marble after a green marble, the total number of marbles in the bag is going to change because event A occurred and removed a marble from the equation. You can use the resources below for extra help.

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