Types of Events - Probability

Simple Event

In probability theory, a simple event is an event that represents a single outcome in the sample space. Here's an example of a simple event:

Suppose we flip a fair coin once. The sample space is {heads, tails}, which has two possible outcomes. A simple event in this scenario would be the event of getting heads, which is a single outcome in the sample space.

Another example of a simple event would be rolling a six-sided die and getting a 3, since that outcome represents a single point in the sample space {1, 2, 3, 4, 5, 6}.

Compound Event

In probability theory, a compound event is an event that consists of two or more simple events. Here's an example of a compound event:

Suppose we roll a six-sided die and flip a fair coin. The sample space for this experiment is {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}, which has 12 possible outcomes. A compound event in this scenario would be the event of rolling an even number and flipping tails. This event consists of two simple events: the event of rolling an even number {2, 4, 6} and the event of flipping tails {T}.

Another example of a compound event would be drawing two cards without replacement from a deck of cards and getting a red card and a face card. This event consists of two simple events: the event of drawing a red card and the event of drawing a face card.

Dependent Event

In probability theory, dependent events are events in which the outcome of one event affects the outcome of another event. Here's an example of dependent events:

Suppose we have a bag with 10 marbles, of which 4 are red and 6 are green. If we draw a marble from the bag without replacement, the probability of drawing a red marble is initially 4/10. However, if we draw a red marble on the first draw and do not replace it, the probability of drawing another red marble on the second draw decreases, since there are now only 3 red marbles remaining in the bag.

To see this, let A be the event of drawing a red marble on the first draw, and let B be the event of drawing a red marble on the second draw. Then, the probability of A is initially 4/10. If A occurs, there are now 3 red marbles and 9 marbles total left in the bag. Therefore, the probability of B given that A has occurred is 3/9.

The probability of both A and B occurring (i.e., drawing two red marbles without replacement) is the probability of A multiplied by the probability of B given that A has occurred:

$$P(A \cap B) = P(A) \cdot P(B|A) = \frac{4}{10} \cdot \frac{3}{9} = \frac{2}{15}$$

Thus, the probability of drawing two red marbles without replacement is smaller than the probability of drawing a red marble on the first draw. This shows that the events are dependent.

Independent Event

In probability theory, independent events are events in which the outcome of one event does not affect the outcome of another event. Here's an example of independent events:

Suppose we flip a fair coin twice. The sample space is {HH, HT, TH, TT}, which has four possible outcomes. The events of getting heads on the first flip and getting heads on the second flip are independent, since the outcome of the first flip does not affect the outcome of the second flip.

To see this, let A be the event of getting heads on the first flip, and let B be the event of getting heads on the second flip. The probability of A is 1/2, and the probability of B given that A has occurred is also 1/2, since the coin is fair and the outcome of one flip does not affect the outcome of another flip. Therefore, the probability of both A and B occurring (i.e., getting heads on both flips) is:

$$P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Thus, the events of getting heads on the first flip and getting heads on the second flip are independent, and the probability of getting heads on both flips is the product of the probabilities of each individual event.

Exhaustive Events

In probability theory, exhaustive events are events that cover all possible outcomes for a given experiment. Here's an example of exhaustive events:

Suppose we roll a six-sided die. The sample space for this experiment is {1, 2, 3, 4, 5, 6}, which has six possible outcomes. The events of getting an even number and getting an odd number are exhaustive, since they cover all possible outcomes.

To see this, let A be the event of getting an even number, and let B be the event of getting an odd number. Then, every number on the die is either even or odd, so every outcome is either in A or in B. Moreover, A and B are mutually exclusive, since no number is both even and odd. Therefore, A and B are exhaustive events.

The probability of getting an even number or an odd number is 1, since every outcome is either even or odd:

$$P(A \cup B) = P(A) + P(B) = \frac{3}{6} + \frac{3}{6} = 1$$

Thus, the events of getting an even number and getting an odd number are exhaustive, and the probability of getting either one is 1.

Mutually Exclusive

Example:

While driving a car, the steering wheel cannot be turned left and right at the same time. Therefore, the events turning left and turning right are considered to be mutually exclusive.

In probability theory, mutually exclusive events are events that cannot occur at the same time. Here's an example of mutually exclusive events:

Suppose we flip a fair coin. The sample space for this experiment is {H, T}, which has two possible outcomes. The events of getting heads and getting tails are mutually exclusive, since the coin can only land on one side.

To see this, let A be the event of getting heads, and let B be the event of getting tails. Then, A and B cannot both occur, since the coin can only land on one side. Therefore, A and B are mutually exclusive events.

The probability of getting either heads or tails is 1, since every outcome is either heads or tails:

$$P(A \cup B) = P(A) + P(B) = \frac{1}{2} + \frac{1}{2} = 1$$

Thus, the events of getting heads and getting tails are mutually exclusive, and the probability of getting either one is 1.

Not Mutually Exclusive

The events are not mutually exclusive: If the events are not mutually exclusive, they may have some outcomes in common. In this case, the probability of the union of the events is the sum of the probabilities of the individual events minus the probability of the intersection of the events.

Example:

Question:

If a 6-sided dice numbered 1 to 6 is rolled, find the probability of getting either an even or prime number.

Solution:

Sample Space={1,2,3,4,5,6}

Let A denote getting an even number.

A = {2,4,6}

Let B denote getting a prime number.

B = {2,3,5}

Let event \(A\) be rolling an even number and event \(B\) be rolling a prime number. Then,

$$A = {2, 4, 6}$$ $$B = {2, 3, 5}$$

$$A \cup B = {2, 3, 4, 5, 6}$$

So, the probability of rolling either an even or prime number is:

$$P(A \cup B) = P({2, 3, 4, 5, 6}) = \frac{5}{6}$$

Therefore, the probability of getting either an even or prime number when rolling a 6-sided dice is \( \frac{5}{6}\) or approximately \(0.83\) .