5  Probability

Understanding probability is integral to understanding research and statistics in psychology. Specficially, probability is a branch of mathematics that deals with the likelihood of an event occurring. It quantifies uncertainty and helps us make informed decisions based on the likelihood of various outcomes. Probability values range from 0 to 1, where 0 indicates an impossible event (i.e., it will never happen), and 1 indicates a certain event (i.e., it’s guaranteed to happen).

5.1 Basic Concepts

The following are some basic concepts we must understand prior to diving into probability. An outcome is a possible result of a situation. An event is A specific set of outcomes. Finally, sample space is the set of all possible outcomes of a situation.

As a practical example, consider flipping a coin. An outcome is a single possible result of the situation; in this case, the outcomes are Heads (H) and Tails (T). The sample space is the set of all possible outcomes, which for a coin flip is represented as {H, T}. Additionally, an event is a specific set of outcomes. For example, we can define Event A as getting Heads, which can be represented as the set {H}. Alternatively, we might define Event B as getting either Heads or Tails, represented as the set {H, T}. This framework helps us understand and calculate probabilities based on different scenarios.

5.2 Calculating Probability

The probability \(P\) of an event \(A\) is calculated using the formula:

\[ P(A) = \frac{\text{Number of favourable outcomes for } A}{\text{Total number of outcomes in sample space}} \]

Tip

Favourable does not mean we necessarily want the outcome. Favourable means that the event in question happens. It’s favourable for the event.

For example, consider a simple experiment: rolling a six-sided die. We want to test to probability of rolling an even number. To this end, we can derive the following:

  • Sample Space (S): {1, 2, 3, 4, 5, 6}
  • Event (A): Rolling an even number {2, 4, 6}

To calculate probability, we can sum the number of favorable outcomes and divide it by the total sample space. There are three favourable outcomes: rolling a 2, 4, or 6. the total sample space is all the possible outcomes, of which there are six: rolling a 1, 2, 3, 4, 5, or 6. Thus:

  1. Number of favorable outcomes for A: 3 (2, 4, 6)
  2. Total number of outcomes in sample space: 6

Using the above formula, we can calculate the probability:

\[ P(A) = \frac{3}{6} = \frac{1}{2} \]

So, the probability of rolling an even number is \(P=\frac{1}{2}=.5\), or 50%.

Practice

Calculate the probability of the following:

  1. A standard six-sided die is rolled. What is the probability of rolling a number greater than 4?

  2. A standard deck of 52 playing cards is shuffled. What is the probability of drawing an Ace from the deck?

  3. A fair coin is flipped once. What is the probability of getting Heads?

  4. A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. If one marble is drawn at random, what is the probability of drawing a blue marble?

  • Favorable outcomes: Rolling a 5 or 6 (2 outcomes: {5, 6})
  • Total outcomes: 6 (numbers: {1, 2, 3, 4, 5, 6})

\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3} \] 2.

  • Favorable outcomes: There are 4 Aces in the deck.
  • Total outcomes: 52 cards

\[ P(A) = \frac{4}{52} = \frac{1}{13} \] 3.

  • Favorable outcomes: Getting Heads (1 outcome: {H})
  • Total outcomes: 2 (Heads and Tails: {H, T})

\[ P(A) = \frac{1}{2} \] 4.

  • Favorable outcomes: There are 3 blue marbles.
  • Total outcomes: 5 red + 3 blue + 2 green = 10 marbles

\[ P(A) = \frac{3}{10} \]

5.3 Compound Events: AND and OR Probabilities

5.3.1 AND Probability

The AND probability (also known as joint probability) is used when we want to find the probability that two events \(A\) and \(B\) both occur. The formula for calculating the AND probability of two independent events is:

\[ P(A \text{ AND } B) = P(A) \times P(B) \]

So, the joint probability can be calculated by multiplying the independent probabilities together.

Consider two independent events:

  • Event A: Rolling a 3 on the die.
  • Event B: Rolling an even number.

We first calculate the probability for each event independently.

  1. Probability of Event A: \(P(A) = \frac{1}{6}\) (only 1 favourable outcome)
  2. Probability of Event B: \(P(B) = \frac{3}{6} = \frac{1}{2}\) (3 favourable outcomes)

Using the formula:

\[ P(A \text{ AND } B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \]

Thus, the probability of rolling a 3 and then rolling an even number is \(\frac{1}{12}\). Note that we cannot roll both a 3 and an even number on one die. Thus, we can think about them as independent events. Rolling a 3 and then rolling an even number. Or, having two dice and having one land on 3 and the other being an even number.

Practice

  1. Two six-sided dice are rolled. What is the probability that both dice show a number greater than 3?

  2. From a standard deck of 52 playing cards, what is the probability of drawing a King and then drawing a Queen without replacement?

  3. A bag contains 4 red marbles and 6 blue marbles. If two marbles are drawn one after the other without replacement, what is the probability that both marbles are red?

  4. Two fair coins are flipped. What is the probability that both coins show Heads?

::: {.callout-tip collapse=“true”} ## Click to Reveal Answers

  • Favorable outcomes: Rolling a 4, 5, or 6 on each die.
  • Total outcomes: Each die has 6 outcomes, so the total outcomes for two dice is \(6 \times 6 = 36\).
  • The probability of one die showing greater than 3 is \(\frac{3}{6} = \frac{1}{2}\).
  • Therefore, for both dice:

\[ P(\text{both > 3}) = P(\text{die 1 > 3}) \times P(\text{die 2 > 3}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. \]

  • Favorable outcomes: 4 Kings, followed by 4 Queens.
  • Total outcomes: 52 cards, then 51 remaining cards. \[ P(\text{King and Queen}) = P(\text{King}) \times P(\text{Queen | King}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663}. \]
  • Favorable outcomes: 4 red marbles.
  • Total outcomes: 10 marbles. \[ P(\text{both red}) = P(\text{first red}) \times P(\text{second red | first red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}. \]
  • Favorable outcomes: Both coins show Heads (1 outcome: {HH}).
  • Total outcomes: 4 outcomes: {HH, HT, TH, TT}. \[ P(\text{both Heads}) = \frac{1}{4}. \] :::

5.3.2 OR Probability

The OR probability is used when we want to find the probability that at least one of two events, \(A\) or \(B\), occurs. The formula for calculating the OR probability is:

\[ P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) \]

Let’s calculate the OR probability for the same events from before. Imagine we have one die and our favourable events are:

  • Event A: Rolling a 3
  • Event B: Rolling an even number.

We can calculate the independent proabilities:

  1. Probability of Event A: \(P(A) = \frac{1}{6}\)
  2. Probability of Event B: \(P(B) = \frac{1}{2}\)
  3. Probability of Event A AND B: \(P(A \text{ AND } B) = 0\) (note: rolling a 3 and an even number is impossible)

We can substitute the values into our formula to derive:

\[ P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) = \frac{1}{6} + \frac{1}{2} - 0 = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3} \]

Thus, the probability of rolling a 3 or an even number is $ $. If we have a fair die, we will makes a bunch of rolls, we would expect to roll a 3 or an even number 75% of the time.

Practice

  1. A standard six-sided die is rolled. What is the probability of rolling a 2 or a 5?

  2. From a standard deck of 52 playing cards, what is the probability of drawing a Heart or a Spade?

  3. In a basket containing 3 apples, 2 oranges, and 5 bananas, what is the probability of randomly selecting an apple or a banana?

  4. A fair coin is flipped twice. What is the probability of getting Heads at least once?

  • Favorable outcomes: Rolling a 2 or a 5 (2 outcomes: {2, 5}).
  • Total outcomes: 6 outcomes. \[ P(2 \text{ or } 5) = P(2) + P(5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}. \]
  • Favorable outcomes: 13 Hearts + 13 Spades = 26.
  • Total outcomes: 52 cards. \[ P(\text{Heart or Spade}) = P(\text{Heart}) + P(\text{Spade}) = \frac{13}{52} + \frac{13}{52} = \frac{26}{52} = \frac{1}{2}. \]
  • Favorable outcomes: 3 apples + 5 bananas = 8.
  • Total outcomes: 10 fruits. \[ P(\text{apple or banana}) = P(\text{apple}) + P(\text{banana}) = \frac{3}{10} + \frac{5}{10} = \frac{8}{10} = \frac{4}{5}. \]
  • Favorable outcomes: HH, HT, TH (3 outcomes).
  • Total outcomes: {HH, HT, TH, TT} (4 outcomes).

\[ P(\text{at least 1 Head}) = \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}=\frac{3}{4} \]

5.4 Conclusion

Understanding probability is crucial in psychological research, particularly in the context of null hypothesis significance testing (NHST; our next chapter). Probability helps researchers determine the likelihood of observing results under the assumption that the null hypothesis is true. By calculating the probability of different outcomes, researchers can assess the strength of their evidence against the null hypothesis, making it easier to identify statistically significant findings. Furthermore, learning how to interpret and combine probabilities allows researchers to better evaluate the risks of Type I and Type II errors, ultimately leading to more informed conclusions and decisions in psychological studies.