A coin is tossed twice. If E and F denote occurrence of head on first toss and second toss respectively, then what is P(E ∪ F) equal to?
Calculating Probability of Coin Toss Events
The problem asks for the probability of getting a head on the first toss or a head on the second toss when a coin is tossed twice. Let's define the sample space and the events involved.
When a fair coin is tossed twice, the possible outcomes form the sample space. Each outcome consists of a sequence of two results (Head or Tail).
- HH: Head on the first toss, Head on the second toss.
- HT: Head on the first toss, Tail on the second toss.
- TH: Tail on the first toss, Head on the second toss.
- TT: Tail on the first toss, Tail on the second toss.
The sample space, denoted by \(S\), is \(\{HH, HT, TH, TT\}\). The total number of possible outcomes is 4. Since the coin is fair, each outcome is equally likely, with a probability of \(\dfrac{1}{4}\).
Let's define the events E and F as given in the question:
- E: Occurrence of head on the first toss. The outcomes corresponding to event E are those where the first result is a Head. So, \(E = \{HH, HT\}\).
- F: Occurrence of head on the second toss. The outcomes corresponding to event F are those where the second result is a Head. So, \(F = \{HH, TH\}\).
Now, let's find the probabilities of events E and F.
- Probability of E, \(P(E) = \dfrac{\text{Number of outcomes in E}}{\text{Total number of outcomes in S}} = \dfrac{2}{4} = \dfrac{1}{2}\).
- Probability of F, \(P(F) = \dfrac{\text{Number of outcomes in F}}{\text{Total number of outcomes in S}} = \dfrac{2}{4} = \dfrac{1}{2}\).
The question asks for the probability of the occurrence of event E or event F. This is the probability of the union of events E and F, denoted as \(P(E \cup F)\). The event \(E \cup F\) includes all outcomes that are in E, or in F, or in both. From our sample space:
- \(E = \{HH, HT\}\)
- \(F = \{HH, TH\}\)
- \(E \cup F = \{HH, HT, TH\}\)
The number of outcomes in \(E \cup F\) is 3. Therefore,
\(P(E \cup F) = \dfrac{\text{Number of outcomes in } E \cup F}{\text{Total number of outcomes in S}} = \dfrac{3}{4}\).
Alternatively, we can use the formula for the probability of the union of two events:
\(P(E \cup F) = P(E) + P(F) - P(E \cap F)\)
Here, \(E \cap F\) is the event where both E and F occur, meaning a head on the first toss AND a head on the second toss. This corresponds to the outcome \(HH\).
- \(E \cap F = \{HH\}\)
- Probability of \(E \cap F\), \(P(E \cap F) = \dfrac{\text{Number of outcomes in } E \cap F}{\text{Total number of outcomes in S}} = \dfrac{1}{4}\).
Now, plug the probabilities into the formula:
\(P(E \cup F) = P(E) + P(F) - P(E \cap F) = \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{4}\)
\(P(E \cup F) = 1 - \dfrac{1}{4} = \dfrac{4}{4} - \dfrac{1}{4} = \dfrac{3}{4}\).
The events E and F are independent events because the result of the first coin toss does not influence the result of the second coin toss. For independent events, the probability of their intersection is the product of their individual probabilities:
\(P(E \cap F) = P(E) \times P(F)\)
Using this property for independent events:
\(P(E \cap F) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).
This confirms our earlier calculation for \(P(E \cap F)\) based on the sample space. Using the union formula for independent events:
\(P(E \cup F) = P(E) + P(F) - P(E)P(F)\)
\(P(E \cup F) = \dfrac{1}{2} + \dfrac{1}{2} - \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}\).
Both methods yield the same result.
Summary of Probabilities
| Event | Outcomes | Number of Outcomes | Probability |
|---|---|---|---|
| Sample Space (S) | {HH, HT, TH, TT} | 4 | 1 |
| E (Head on first toss) | {HH, HT} | 2 | \(\dfrac{2}{4} = \dfrac{1}{2}\) |
| F (Head on second toss) | {HH, TH} | 2 | \(\dfrac{2}{4} = \dfrac{1}{2}\) |
| \(E \cap F\) (Head on both tosses) | {HH} | 1 | \(\dfrac{1}{4}\) |
| \(E \cup F\) (Head on first OR second toss) | {HH, HT, TH} | 3 | \(\dfrac{3}{4}\) |
The probability \(P(E \cup F)\) is therefore \(\dfrac{3}{4}\).
Revision Table: Key Probability Concepts
| Concept | Description | Formula Example |
|---|---|---|
| Sample Space (S) | Set of all possible outcomes of an experiment. | For tossing a coin twice: S = {HH, HT, TH, TT} |
| Event | A subset of the sample space. | E = {HH, HT} (Head on first toss) |
| Probability of an Event A | \(P(A) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\) (for equally likely outcomes) | \(P(E) = \dfrac{2}{4} = \dfrac{1}{2}\) |
| Intersection of Events (\(A \cap B\)) | Event where both A and B occur. | \(E \cap F = \{HH\}\) |
| Union of Events (\(A \cup B\)) | Event where A occurs OR B occurs OR both occur. | \(E \cup F = \{HH, HT, TH\}\) |
| Probability of Union | \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) | \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\) |
| Independent Events | The occurrence of one event does not affect the probability of the other. | Coin tosses are typically independent. \(P(A \cap B) = P(A)P(B)\) if A and B are independent. |
Additional Information: Probability of Union and Independent Events
Understanding the probability of the union of events is crucial in probability theory. The formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) is a fundamental rule, often called the Addition Rule. The term \(P(A \cap B)\) is subtracted because the outcomes where both A and B occur are counted twice (once in P(A) and once in P(B)).
When events A and B are mutually exclusive (meaning they cannot happen at the same time, so \(A \cap B = \emptyset\)), then \(P(A \cap B) = 0\). In this special case, the Addition Rule simplifies to \(P(A \cup B) = P(A) + P(B)\). However, in our coin toss example, events E (Head on first) and F (Head on second) are NOT mutually exclusive because the outcome HH is in both events.
The concept of independent events is also very important. Two events A and B are independent if \(P(A \cap B) = P(A) \times P(B)\). The outcome of the first coin toss does not change the probabilities of the outcomes of the second coin toss, making E and F independent. This property allows us to calculate \(P(E \cap F)\) easily if we know \(P(E)\) and \(P(F)\).
For independent events E and F, the Addition Rule can also be written as \(P(E \cup F) = P(E) + P(F) - P(E)P(F)\), which we used and verified in the solution.
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