In Target A, we talked about how to determine the possible number of outcomes in a given scenario. We are now going to look at finding the probability of a certain outcome occurring. You have done this before with simple scenarios. For example, if you wanted to know the probability of rolling a number greater than \(4\) on a six-sided die, you would first determine the total number of possible outcomes which is \(6\) and then the desired number of outcomes which would be \(2\) since rolling either a \(5\) or a \(6\) would meet the condition of being greater than \(4\). We would then say our probability is \(\large\frac{2}{6}=\frac{1}{3}\), and we can also write this as \(0.333\) or \(33.3\%\).
The probability of event A happening is given by the formula \(P\left(A\right)=\large\frac{\text{Number of outcomes in event A}}{\text{Total number of outcomes}}\).
The probability of an event is the likelihood the event occurs. Probabilities will always be numbers between \(0\) (or \(0\%\)) and \(1\) (or \(100\%\)). This is because the fewest number of desired outcomes is \(0\) and the most would be equivalent to the total number of outcomes which would give you a probability of \(1\).
We’ll work often with problems involving a standard deck of fifty-two playing cards. Here is an image to help you if you need to learn or refresh your memory on the types of cards in a deck.
The probability of event A happening is given by the formula \(P\left(A\right)=\large\frac{\text{Number of outcomes in event A}}{\text{Total number of outcomes}}\).
The probability of an event is the likelihood the event occurs. Probabilities will always be numbers between \(0\) (or \(0\%\)) and \(1\) (or \(100\%\)). This is because the fewest number of desired outcomes is \(0\) and the most would be equivalent to the total number of outcomes which would give you a probability of \(1\).
We’ll work often with problems involving a standard deck of fifty-two playing cards. Here is an image to help you if you need to learn or refresh your memory on the types of cards in a deck.
Example 1: You pick one card from a standard deck of \(52\) playing cards. Find each probability.
a) \(P\left(7\right)\)
There are four \(7\)’s in the deck so \(4\) would be our number of desired outcomes, and our total number is \(52\).
\(P\left(7\right)=\frac{4}{52}=\frac{1}{13}\ \text{or}\ 0.077\ \text{or}\ 7.7\%\)
b) \(P\left(\text{a card other than a K}\right)\)
There are \(48\) cards that are not a King so that would be our number of desired outcomes.
\(P\left(\text{a card other than a K}\right)=\large\frac{48}{52}=\frac{12}{13}\ \normalsize\text{or}\ 0.923\ \text{or}\ 92.3%\)
Example 2: You win a certain lottery if you correctly select \(5\) of the \(70\) numbers. The order of the number is not important, and you can only pick each number once. Find the probability that you will win the lottery.
We talked about a jury setup when we first introduced combinations. If you were asked to find the probability of selecting a female for a given juror spot, you would just take the total number of females and divide it by the total number of people in the jury pool. This gets a bit more complicated if we want to find the probability of a jury being composed of, let’s say, \(7\) men and \(5\) women. Since the order of juror selection doesn’t matter, we need to use combinations to figure out the total number of possible outcomes that includes \(7\) men and \(5\) women.
Example 3: A jury of \(12\) people is selected from a pool of \(40\) people that includes \(23\) men and \(17\) women. Find the probability that the jury will be composed of \(7\) men and \(5\) women.
The probability of this occurring would come from \(\large\frac{\text{Total number of juries from this pool composed of 7 men and 5 women}}{\text{Total number of juries that can be made from this pool}}\).
We use combinations to help us do this.
\(\large\frac{_{23}C_7\cdot_{17}C_5}{_{40}C_{12}}=\frac{1,517,031,516}{5,586,853,480}\ \normalsize\text{or}\ 27.15\%\)
Now let’s say we want to find the probability of the jury being composed of no more than \(2\) men.
Example 4: You are going to take four books from your friend’s library on vacation. There are a total of \(30\) books – five are biographies, seventeen are fiction books, two are travel books, and six are nonfiction books. Assume the books are chosen at random.
a) What is the probability that exactly \(3\) of the books will be fiction?
b) What is the probability that you will take one of each type of book?
This scenario only works if we select one biography and one fiction and one travel and one nonfiction. This forms the setup for the numerator.
\(\large\frac{_5C_1\cdot_{17}C_1\cdot_2C_1\cdot_6C_1}{_{30}C_4}=\frac{1020}{27405}=\frac{68}{1827}\ \normalsize\text{or}\ 3.72\%\)
Probability of Compound Events
Probability of Independent Events
Probability of Dependent Events
Guided Learning Extension-Expected Value