Additional law of probability

Addition Law of Probability

The addition law of probability is used to determine the probability that at least one of multiple events will occur.

Key Concepts

• Events: An event is a specific outcome or a set of outcomes of a random experiment.
• Union of Events ($A \cup B$): The event that either $A$ or $B$ or both occur.
• Intersection of Events ($A \cap B$): The event that both $A$ and $B$ occur simultaneously.
• Mutually Exclusive Events: Two events that cannot happen at the same time.

General Addition Law:

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

This formula accounts for the overlap (intersection) where both events occur, ensuring it is not double-counted.

Example(Non-Mutually Exclusive Events):

Suppose event $A$ is drawing a red card from a deck of cards $P(A) = \frac{26}{52} = \frac{1}{2}$, and event $B$ is drawing a king from the deck $P(B) = \frac{4}{52} = \frac{1}{13}$. The probability of drawing a red king (both events occurring) $P(A \cap B)$ is $\frac{2}{52} = \frac{1}{26}$.

So,

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{13} - \frac{1}{26} = \frac{15}{26}$

Formula for Mutually Exclusive Events:

If $A$ and $B$ are mutually exclusive (cannot happen at the same time), $P(A \cap B) = 0$. So,

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

Example:

Suppose the probability of event $A$ (rolling a 3 on a die) is $\frac{1}{6}$, and the probability of event $B$ (rolling a 4 on a die) is $\frac{1}{6}$. Since these events are mutually exclusive:

So,

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

Practice Question

Question 1. Two unbiased ice are thrown find the probability that the sum of numbers obtained on the two dice is neither a multple of 2 nor a multiple of 3.

Solution:

Let S be the sample space; In throwing of 2 dice the total number of outcomes i.e.,

$n(S) = 36$

Let A be the event such that sum of two numbers is a multiple of 2, and B be the event such that the sum of two numbers is a multiple of 3. Then,

A = Sum is a multpile of 2 = $\begin{pmatrix} (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), \\ (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), \\ (5,1), (5,3), (5,5), (6,2), (6,4), (6,6) \end{pmatrix}$

$n(A) = 18 \implies P(A) = \frac{18}{36} = \frac{1}{2}$

Also, B = Sum is a multiple of 3 = $\begin{pmatrix} (1,2), (1,5), (2,1), (2,4), (3,3), (3,6) \\ (4,2), (4,5), (5,1), (5,4), (6,3), (6,6) \end{pmatrix}$

$n(B) = 12 \implies P(B) = \frac{12}{36} = \frac{1}{3}$

Also, $A \cap B = \{ (1,5), (2,4), (3,3), (4,2), (5,1), (6,6) \}$

$n(A \cap B) = 6 \implies P(A \cap B) = \frac{6}{36} = \frac{1}{6}$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3}$

Therefore, the required probability $= 1 - P(A \cup B) = 1 - \frac{2}{3} = \frac{1}{3}$

Question 2. if A and B are two events such that $P(A) = \frac{1}{4}, P(B)=\frac{1}{2} and P(A \cap B) = \frac{1}{8}$
Find (i)P(A or B)     (ii)P(not A and not B)

Solution:

Given that,
$P(A) = \frac{1}{4}, \quad P(B) = \frac{1}{2} \quad \text{and} \quad P(A \cap B) = \frac{1}{8}$
(i) By addition theorem, we have
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\implies P(A \cup B) = \frac{1}{4}+\frac{1}{2}=\frac{1}{8}$
i.e.,
$P(A \text{ or } B) = \frac{5}{8}$
(ii)$P(\text{not } A \text{ and not } B) = P(\overline{A} \cap \overline{B})$
$\implies P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B})$
$\implies P(\overline{A \cup B}) = 1 - P(A \cup B)$
$\implies 1 - \frac{5}{8} = \frac{3}{8}$