Probability of an Event and Addition Rules

1. Probability of an Event (Equally Likely Outcomes)

Based on the axiomatic approach, we can derive the classical formula for probability when all outcomes in a sample space are equally likely.

Let $S$ be a sample space with $n(S)$ total outcomes, all of which are equally likely. Let E be an event with $n(E)$ favorable outcomes. The probability of event E occurring is given by:

$$P(E) = \frac{\text{Number of outcomes favorable to E}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$$

• Example: A card is drawn from a standard deck of 52 cards. What is the probability of drawing a King?
  • Total outcomes $n(S) = 52$.
  • Favorable outcomes (number of Kings) $n(E) = 4$.
  • Probability $P(E) = \frac{4}{52} = \frac{1}{13}$.

2. Addition Rule of Probability

The axioms of probability can be extended to find the probability of the union of two events, even when they are not mutually exclusive.

A. For any Two Events

For any two events A and B in a sample space S, the probability of 'A or B' (i.e., the probability that at least one of the events occurs) is given by:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

We subtract $P(A \cap B)$ because this portion was counted twice: once in $P(A)$ and once in $P(B)$.

B. For Mutually Exclusive Events

If A and B are mutually exclusive events, then they have no outcomes in common, so $A \cap B = \emptyset$. This means $P(A \cap B) = P(\emptyset) = 0$.

In this special case, the addition rule simplifies to:

$$P(A \cup B) = P(A) + P(B)$$

This is a direct restatement of the third axiom of probability.

3. Addition Rule for Three Events

The rule can be extended to three events A, B, and C:

$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$

Solved Examples

Example 1: Simple Probability

Question: Two fair coins are tossed. What is the probability of getting at least one head?

Explanation:

1. Define the Sample Space (S): The sample space is the set of all possible outcomes. When tossing two coins, the possibilities are: $S = \{HH, HT, TH, TT\}$. The total number of outcomes is $n(S) = 4$.

2. Define the Event (E): The event is 'getting at least one head'. The outcomes favorable to this event are: $E = \{HH, HT, TH\}$. The number of favorable outcomes is $n(E) = 3$.

3. Calculate the Probability: The probability of an event is the ratio of favorable outcomes to the total number of outcomes. $P(E) = \frac{n(E)}{n(S)} = \frac{3}{4}$.

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Example 2: Using the Complement

Question: Using the previous example, find the probability of getting at least one head by using the complement rule.

Explanation: Sometimes, calculating the probability of the event you don't want is easier.

1. Define the Event (E) and its Complement (E'):

   • The event is E = 'getting at least one head'.
   • The complementary event, E', is the opposite: 'not getting at least one head', which means 'getting no heads'.

2. Calculate the Probability of the Complement: The only outcome where there are no heads is $E' = \{TT\}$. The probability of the complement is $P(E') = \frac{n(E')}{n(S)} = \frac{1}{4}$.

3. Apply the Complement Rule: The probability of an event is 1 minus the probability of its complement: $P(E) = 1 - P(E')$. $P(E) = 1 - \frac{1}{4} = \frac{3}{4}$.

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Example 3: Addition Rule

Question: A card is drawn from a deck of 52 cards. What is the probability that the card is either a spade or a king?

Explanation:

1. Define the Events:

   • A = 'the card is a spade'.
   • B = 'the card is a king'. We need to find $P(A \cup B)$.

2. Check for Mutual Exclusivity: These events are not mutually exclusive because there is an outcome in common: the King of Spades.

3. Use the General Addition Rule: For non-mutually exclusive events, the formula is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

4. Calculate Each Probability:

   • $P(A) = \frac{13}{52}$ (since there are 13 spades).
   • $P(B) = \frac{4}{52}$ (since there are 4 kings).
   • $P(A \cap B)$ (the probability of being the King of Spades) = $\frac{1}{52}$.

5. Solve: $P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{17 - 1}{52} = \frac{16}{52} = \frac{4}{13}$.

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Example 4: Mutually Exclusive Events

Question: A single die is rolled. What is the probability of getting a 2 or a 5?

Explanation:

1. Define the Events:

   • A = 'getting a 2'.
   • B = 'getting a 5'.

2. Check for Mutual Exclusivity: A single die roll cannot be both a 2 and a 5. Therefore, the events are mutually exclusive, and their intersection is empty ($A \cap B = \emptyset$).

3. Use the Simplified Addition Rule: For mutually exclusive events, $P(A \cup B) = P(A) + P(B)$.

4. Calculate and Solve: $P(A) = \frac{1}{6}$ and $P(B) = \frac{1}{6}$. $P(A \cup B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

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Example 5: Finding Intersection Probability

Question: If $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \cup B) = 0.7$, what is $P(A \cap B)$?

Explanation:

1. Start with the General Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

2. Rearrange the Formula: We can algebraically rearrange the formula to solve for the intersection: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$.

3. Substitute and Solve: $P(A \cap B) = 0.5 + 0.4 - 0.7 = 0.9 - 0.7 = 0.2$.

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Example 6: Probability of 'A but not B'

Question: In a class of 30 students, 20 like football (F) and 15 like cricket (C). 10 students like both. What is the probability that a randomly selected student likes football but not cricket?

Explanation:

1. Identify the Goal: We want to find the probability of the event 'F but not C', which is written as $P(F - C)$.

2. Use the Formula: The formula for this is $P(F - C) = P(F) - P(F \cap C)$. This formula finds the probability of F and subtracts the overlapping part with C.

3. Find the Probabilities from the Data:

   • $P(F) = \frac{\text{Number who like football}}{\text{Total students}} = \frac{20}{30}$.
   • $P(F \cap C) = \frac{\text{Number who like both}}{\text{Total students}} = \frac{10}{30}$.

4. Calculate the Result: $P(F - C) = \frac{20}{30} - \frac{10}{30} = \frac{10}{30} = \frac{1}{3}$.

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Example 7: Addition Rule for Three Events

Question: A survey shows that 63% of people like cheese (C), 76% like apples (A), and 55% like bananas (B). Also, $P(C \cap A) = 40\%$, $P(C \cap B) = 35\%$, $P(A \cap B) = 45\%$, and $P(A \cap B \cap C) = 25\%$. What is the probability that a person likes at least one of the three?

Explanation:

1. Identify the Goal: We need to find the probability of the union of the three events, $P(A \cup B \cup C)$.

2. Use the Inclusion-Exclusion Principle for Three Events: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$.

3. Substitute the Given Probabilities (in decimal form): $P(A \cup B \cup C) = 0.76 + 0.55 + 0.63 - 0.45 - 0.35 - 0.40 + 0.25$.

4. Calculate:

   • Sum of individual probabilities: $0.76 + 0.55 + 0.63 = 1.94$.
   • Sum of pairwise intersections: $0.45 + 0.35 + 0.40 = 1.20$.
   • Triple intersection: $0.25$. $P(A \cup B \cup C) = 1.94 - 1.20 + 0.25 = 0.74 + 0.25 = 0.99$.

   So, there is a 99% probability that a randomly selected person likes at least one of the three items.

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Example 8: Probability with Combinations

Question: A committee of 2 is to be selected from 2 men and 2 women. What is the probability that the committee will have one man?

Explanation:

1. Calculate the Total Number of Outcomes (Sample Space): There are 4 people in total, and we are selecting a committee of 2. The order does not matter, so we use combinations. $n(S) = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$.

2. Calculate the Number of Favorable Outcomes: The event E is 'selecting one man'. To form the committee, we must select 1 man from the 2 available men AND 1 woman from the 2 available women.

   • Ways to choose 1 man from 2: $\binom{2}{1} = 2$.
   • Ways to choose 1 woman from 2: $\binom{2}{1} = 2$. The total number of favorable outcomes is $n(E) = \binom{2}{1} \times \binom{2}{1} = 2 \times 2 = 4$.

3. Calculate the Probability: $P(E) = \frac{n(E)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.

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Example 9: Using Complement and 'At least one'

Question: Four people are asked to pick a random number from 1 to 10. What is the probability that at least two people pick the same number?

Explanation:

1. Identify the Best Strategy: Calculating the probability of 'at least two' (which could be exactly 2, 3, or 4) is complicated. It's much easier to calculate the probability of the complementary event and subtract from 1.

   • E = 'at least two people pick the same number'.
   • E' = 'no two people pick the same number' (i.e., everyone picks a different number).

2. Calculate Total Outcomes, n(S): Each of the 4 people has 10 independent choices. So, $n(S) = 10 \times 10 \times 10 \times 10 = 10^4 = 10000$.

3. Calculate Favorable Outcomes for the Complement, n(E'): For everyone to pick a different number:

   • The first person has 10 choices.
   • The second person has 9 remaining choices.
   • The third has 8 remaining choices.
   • The fourth has 7 remaining choices. $n(E') = 10 \times 9 \times 8 \times 7 = 5040$.

4. Calculate the Probability of the Complement: $P(E') = \frac{n(E')}{n(S)} = \frac{5040}{10000} = 0.504$.

5. Calculate the Final Probability: $P(E) = 1 - P(E') = 1 - 0.504 = 0.496$.

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Example 10: Finding $P(A' \cap B')$

Question: Given $P(A) = 0.4$, $P(B)=0.5$ and $P(A \cup B) = 0.6$. Find the probability that neither A nor B occurs.

Explanation:

1. Translate the Goal: The event 'neither A nor B occurs' is the intersection of the complements: $A' \cap B'$.

2. Apply De Morgan's Law: De Morgan's Law states that the intersection of the complements is the same as the complement of the union: $A' \cap B' = (A \cup B)'$. So, we need to find $P((A \cup B)')$.

3. Use the Complement Rule: The probability of the complement of an event is 1 minus the probability of the event itself. $P((A \cup B)') = 1 - P(A \cup B)$.

4. Substitute and Solve: We are given $P(A \cup B) = 0.6$. $P(A' \cap B') = 1 - 0.6 = 0.4$.