Probability Rules(With Formulas & Examples)

Probability is a measure of the likelihood of an event occurring. Many events cannot be predicted with total certainty. Using probability, one can predict only the chance of an event to occur, i.e., how likely they are going to happen.

There are 5 basic probability rules that are used to solve problems. Let’s understand these rules in probability with examples.

6 Basic Probability Rules

The six basic rules in probability are as follows.

Rule 1: The probability of an impossible event is $0$; the probability of a certain event is $1$. Therefore, for any event $\text{A}$, the range of possible probabilities is $0 \le \text{P}( \text{A} ) \le 1$.

Rule 2: For $\text{S}$ the sample space of all possibilities, $\text{P} (\text{S}) = 1$, i.e., the sum of all the probabilities for all possible events is equal to one. 

Rule 3: For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$. It follows then that $\text{P}(\text{A}) = 1 – \text{P}( \text{A}^{\text{c}})$

Rule 4 (Addition Rule): This probability rule relates to either one or both events that occur

• If two events say $\text{A}$ and $\text{B}$, are mutually exclusive i.e., $\text{A}$ and $\text{B}$ have no outcomes in common then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B})$
• If two events are not mutually exclusive, then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A and B})$

Rule 5 (Conditional Probability): $\text{P}(\text{A} | \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} | \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$.

Note: The straight line symbol, $|$, does not mean divide! This symbol means “conditional” or “given”. For example $\text{P}(\text{A}|\text{B})$ means the probability that event $\text{A}$ occurs given event $\text{B}$ has already occurred.

Rule 6 (Multiplication Rule): This probability rule is used when both events occur

$\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}| \text{B})$

If $\text{A}$ and $\text{B}$ are independent, i.e., neither event influences or affects the probability of other events then $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B})$. This particular rule extends to more than two independent events. In that case, it becomes, $\text{P}(\text{A and B and C}) = \text{P}(\text{A}) \times \text{P}(\text{B}) \times \text{P}(\text{C})$.

Rule 1: $0 \le \text{P}( \text{A} ) \le 1$

To prove the statement that for any event $\text{A}$, $0 \le \text{P}( \text{A} ) \le 1$, we’ll use the following three axioms:

1. The probability of an event will always be greater than or equal to zero i.e. $\text{P}(\text{A}) \ge 0$ for any event $\text{A}$. (Since the probability of an impossible event is $0$ and the probability of an event cannot be negative).
2. The probability of a sample space will always be equal to $1$ i.e. $\text{P}(\text{S}) = 1$.
3. Given some mutually exclusive events, the probability of the union of all these mutually exclusive events will always be equal to the summation of the probability of individual events i.e. $\text{P} \left(\text{A}_1 \cup \text{A}_2 \cup \text{A}_3 \cup \text{A}_4 … \cup \text{A}_{\text{n}} \right) =\text{P} \left(\text{A}_1 \right) + \text{P} \left(\text{A}_2 \right) + \text{P} \left(\text{A}_3 \right) + \text{P} \left(\text{A}_4 \right) + … + \text{P} \left(\text{A}_n \right)$ 

According to axiom 1, the probability of an event will always be greater than or equal to $0$.

$\text{P} \left( \text{A} \right) \ge 0$ (According to Axiom 1)  — (1)

The probability of a sample space will be equal to the probability of the intersection of $\text{A}$ and $(\text{S} – \text{A})$ i.e.

$\text{S} = \text{A} + (\text{S} – \text{A})$

$\text{P}(\text{S}) = \text{P}(\text{A} + (\text{S} – \text{A}))$

Since $\text{A}$ and $(\text{S} – \text{A})$ are two mutually exclusive events. So, according to axiom 3, it can be written as $\text{P}(\text{A} + (\text{S} – \text{A})) = \text{P}(\text{A}) + \text{P}(\text{S} – \text{A})$

This implies, $\text{P}(\text{S}) = \text{P}(\text{A}) + \text{P}(\text{S} – \text{A}))$— (2)

Now, from axiom 1, it can be said that the $\text{P}(\text{S} – \text{A})$ will always be greater than or equal to zero i.e. $\text{P}(\text{S} – \text{A}) \ge 0$.

If something positive is added to a given value, its value will always increase. Since, $\text{P}(\text{S} – \text{A}) \ge 0$, it can be said that $\text{P}(\text{A})$ can’t be greater than $\text{P}(\text{S})$. Otherwise, equation (2) will not hold true. 

This means $\text{P}(\text{S}) \ge \text{P}(\text{A})$

From axiom 2, the probability of a Sample Space always equals 1. So, this means $1 \ge \text{P}(\text{A})$,  or $\text{P} (\text{A}) \ge 1$ — (3)

From equations (1) and (3), it can be shown that $0 \le \text{P}(\text{A}) \le 1$

This proves that the probability of an event will always lie between $0$ and $1$.

Examples of $0 \le \text{P}( \text{A} ) \le 1$

Example 1: Find the probability of getting a sum of $15$ in a throw of a pair of dice.

Let $\text{E}$ be the event of getting a sum of $15$ in a throw of a pair of dice.

The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$

$ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), $

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), $

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), $

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), $

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$.

From the above sample space, we see that there is no pair of numbers where the sum is $15$, therefore, the total number of favourable events is $m = 0$.

Therefore, the probability of  getting a sum of $15$ in a throw of a pair of dice $\text{P}(\text{E}) = \frac{m}{n} = \frac{0}{36} = 0$.

Example 2: Find the probability of getting a positive number less than $13$ as a sum in a throw of a pair of dice.

Let $\text{E}$ be the event of getting a positive number less than $13$ as a sum in a throw of a pair of dice.

The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$

$ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), $

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), $

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), $

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$.

From the above sample space, we see that for all the pairs the sum is a positive number less than $13$,  therefore, the total number of favourable events is $m = 36$.

Therefore, the probability of getting a positive number less than $13$ as a sum in a throw of a pair of dice $\text{P}(\text{E}) = \frac{m}{n} = \frac{36}{36} = 1$.

Example 3: Find the probability of picking a non-face card from a deck of well-shuffled cards.

The number of cards in a deck of cards $n = 52$.

The number of face cards in a deck of cards = $12$.

Therefore, the number of non-face cards in a deck of cards $m = 52 – 12 = 40$

Thus, the probability of picking a non-face card from a deck of a well-shuffled pack of $52$ cards is $\frac{m}{n} = \frac{40}{52} = \frac{10}{13} = 0.7692$.

Note: From the above three examples, we see that $0 \le \text{P}( \text{A} ) \le 1$.

Rule 2: $\text{P} (\text{S}) = 1$

Let $\text{S}$ be a sample space consisting of events $\text{E}_1$, $\text{E}_2$, $\text{E}_3$, $\text{E}_4$, …, $\text{E}_n$, then $\text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup … \text{E}_n = \text{S}$ ————– (1)

We know that probability of an event $\text{E}$ is given by $\text{P}(\text{E}) = \frac{n(\text{E})}{n(\text{S})}$, therefore, $\text{P}(\text{E}_1) + \text{P}(\text{E}_2) + \text{P}(\text{E}_3) + \text{P}(\text{E}_4) + … + \text{P}(\text{E}_n) = \frac{n(\text{E}_1)}{n(\text{S})} + \frac{n(\text{E}_2)}{n(\text{S})} + \frac{n(\text{E}_3)}{n(\text{S})} + \frac{n(\text{E}_4)}{n(\text{S})} + \frac{n(\text{E}_n)}{n(\text{S})}$

$= \frac{n(\text{E}_1) + n(\text{E}_2) + n(\text{E}_3) + n(\text{E}_4) + … + n(\text{E}_n)}{n(\text{S})}$

$= \frac{n \left(\text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup … \text{E}_n \right)}{n(\text{S})}$

$= \frac{n(\text{S})}{n(\text{S})} = 1$

Examples of $\text{P} (\text{S}) = 1$

Example 1: Consider a random experiment of tossing a pair of coins.

The sample space is given by $\text{S} = \{(\text{H}, \text{H}), (\text{H}, \text{T}), (\text{T}, \text{H}), (\text{T}, \text{T}) \}$.

Probability of an event $(\text{H}, \text{H}) = \frac{1}{4}$

Probability of an event $(\text{H}, \text{T}) = \frac{1}{4}$

Probability of an event $(\text{T}, \text{H}) = \frac{1}{4}$

Probability of an event $(\text{T}, \text{T}) = \frac{1}{4}$

Therefore, $\text{P}(\text{H}, \text{H}) + \text{P}(\text{H}, \text{T}) + \text{P}(\text{T}, \text{H}) + \text{P}(\text{T}, \text{T})$

$ = \text{P}((\text{H}, \text{H}) + (\text{H}, \text{T}) + (\text{T}, \text{H}) + (\text{T}, \text{T})) = \text{P}(\text{S})$

$ = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1$

Rule 3: Complementary Events Rule

For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$

For any event, $\text{A}$, the complementary event $\text{A}^{\text{c}}$ is an event that includes all the events of a sample space except event $\text{A}$.

Therefore, if $\text{S}$ is a sample space such that $\text{S} = \text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup … \text{E}_n$, then complement of event $\text{E}_1$, written as ${\text{E}_{1}}^{c}$ is given by ${\text{E}_{1}}^{c} = \text{S} – \text{E}_1$

Thus, the probability of complementary event ${\text{E}_{1}}^{c}$ is given by $\text{P} \left(\text{S} – \text{E}_1 \right)$

$=>\text{P} \left({\text{E}_{1}}^{c} \right) = \text{P} \left(\text{S} \right) – \text{P}(\text{E}_1 )$

$=>\text{P} \left({\text{E}_{1}}^{c} \right) = 1 – \text{P}(\text{E}_1 )$

Thus, For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$.

Examples of Complementary Events

Example 1: A number is chosen at random from a set of whole numbers from 1 to 50. Calculate the probability that the chosen number is not a perfect square.

Number of numbers from $1$ to $50$, $n = 50$.

Perfect squares from $1$ to $50$ are $1$, $4$, $9$, $16$, $25$, $36$, and $49$.

Therefore, number of perfect squares from $1$ to $50$, $m = 7$

Let $\text{E}$ be the event of choosing a perfect square number, therefore, $\text{P}(\text{E}) = \frac{m}{n} = \frac{7}{50}$.

Therefore, the probability that the chosen number is not a perfect square, $\text{P}(\text{E}^{\text{c}}) = 1 – \text{P}(\text{E}) = 1 – \frac{7}{50} = \frac{43}{50}$.

Example 2: There are $10$ balls in a bag out of which $3$ are black, $2$ are red, $1$ is blue, $2$ are pink, and $2$ are purple. Find the probability of not picking a pink ball.

The number of balls in a bag $n = 3 + 2 + 1 + 2 + 2 = 10$.

The number of pink balls in a bag $m = 2$.

Let $\text{E}$ be the event of picking a pink ball.

Therefore, the probability of picking a pink ball $\text{P}(\text{E}) = \frac{m}{n} = \frac{2}{10} = \frac{1}{5}$.

Thus, the probability of not picking a pink ball = $1 – \text{P}(\text{E}) = 1 – \frac{1}{5} = \frac{4}{5}$.

Example 3: A card is chosen from a deck of well-shuffled cards. What is the probability that the number on the card is not $4$?

The number of cards in a deck of cards $n = 52$.

The number of cards having a number 2 $m = 4$.

Let $\text{E}$ be the event of picking a card with the number $2$ on it.

Therefore, the probability of picking a card with number $2$ on it $\text{P}(\text{E}) = \frac{m}{n} = \frac{4}{52} = \frac{1}{13}$.

Thus, the probability of not card with number $2$ = $1 – \text{P}(\text{E}) = 1 – \frac{1}{13} = \frac{12}{13}$.

Rule 4: Addition Rule of Probability

Let $\text{A}$ and $\text{B}$ be two events.

Case 1: When $\text{A}$ and $\text{B}$ are non-mutually exclusive events(there exist a common event between the two events), then  $\text{A} \cup \text{B} = \text{A} + \text{B} – \text{A} \cap \text{B}$

Therefore, $\text{P}(\text{A} \cup \text{B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A} \cap \text{B})$

Case 2: When $\text{A}$ and $\text{B}$ are mutually exclusive events(there doesnot exist a common event between the two events), then  $\text{A} \cup \text{B} = \text{A} + \text{B}$

Therefore, $\text{P}(\text{A} \cup \text{B}) = \text{P}(\text{A}) + \text{P}(\text{B})$

Examples of Addition Rule of Probability

Example 1: There are $50$ cards numbered from $1$ to $50$. A card is chosen at random. What is the probability that the number on the card is either less than $10$ or greater than $40$.

The total number of cards $n = 50$.

Let $\text{E}_1$ be an event of picking a card with a number less than $10$.

And, Let $\text{E}_2$ be an event of picking a card with a number greater than $40$.

Numbers less than $10$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, and $9$.

Therefore, a number of numbers less than $10$ is $9$.

Numbers greater than $40$ are $41$, $42$, $43$, $44$, $45$, $46$, $47$, $48$, $49$, and $50$.

Therefore, a number of numbers greater than $40$ are $10$.

Thus, $\text{P}\left(\text{E}_1 \right) = \frac{9}{50}$ and $\text{P}\left(\text{E}_2 \right) = \frac{10}{50}$

Hence, the probability that the number on the card is either less than $10$ or greater than $40$ is $\text{P}\left(\text{E}_1 \right) + \text{P}\left(\text{E}_2 \right) = \frac{9}{50} + \frac{10}{50} = \frac{19}{50}$.

Note: In this example, the events number on the card are either less than $10$ or greater than $40$ are non-mutually exclusive events.

Example 2: There are $50$ cards numbered from $1$ to $50$. A card is chosen at random. What is the probability that the number on the card is either divisible by $3$ or divisible by $5$?

The total number of cards $n = 50$.

Let $\text{E}_1$ be an event of picking a card with a number divisible by $3$.

And, Let $\text{E}_2$ be an event of picking a card with a number divisible by $5$.

Numbers divisible by $3$ are $3$, $6$, $9$, $12$, $15$, $18$, $21$, $24$, $27$, $30$, $33$, $36$, $39$, $42$, $45$, and $48$.

Therefore, the number of numbers divisible by $3$ is $16$.

Numbers divisible by $5$ are $5$, $10$, $15$, $20$, $25$, $30$, $35$, $40$, $45$, and $50$.

Therefore, the number of numbers divisible by $5$ is $10$.

Numbers divisible by both $3$ and $5$ are $15$, $30$, and $45$.

Therefore, the number of numbers divisible by both $3$ and $5$ is $3$.

Let $\text{E}_1$ be an event of picking a card with a number divisible by $3$.

and  $\text{E}_2$ be an event of picking a card with a number divisible by $5$.

Therefore, $\text{E}_1 \cap \text{E}_2$ will be an event of picking a card divisible by both $3$ and $5$.

Thus, the probability that the number on the card is either divisible by $3$ or divisible by $5$ = $\text{P}(\text{E}_1) + \text{P}(\text{E}_2) – \text{P}(\text{E}_1 \cap \text{E}_2)$

$= \frac{16}{50} + \frac{10}{50} – \frac{3}{50} = \frac{23}{50}$.

Note: In this example, the events number on the card are either multiple of $3$ or multiple of $5$ are mutually exclusive events.

Rule 5: Conditional Probability

The conditional probability rule states that If $\text{A}$ and $\text{B}$ are two events that $\text{P}(\text{A} | \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} | \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$.

When the intersection of two events happens, then the formula for conditional probability for the occurrence of two events is given by $\text{P}(\text{A}|\text{B}) = \frac{n(\text{A} \cap \text{B})}{n(\text{B})}$, or $\text{P}(\text{B}| \text{A}) = \frac{n(\text{A} \cap \text{B})}{n(\text{A})}$

where $\text{P}(\text{A}| \text{B})$ represents the probability of occurrence of $\text{A}$ given $\text{B}$ has occurred.

$n(\text{A} \cap \text{B})$ is the number of elements common to both $\text{A}$ and $\text{B}$.

$n(\text{B})$ is the number of elements in $\text{B}$, and it cannot be equal to zero.

Let $n$ represent the total number of elements in the sample space, then $\text{P}(\text{A}| \text{B}) = \frac{\frac{n(\text{A} \cap \text{B})}{n}}{\frac{n(\text{B})}{n}}$

Since $\frac{n(\text{A} \cap \text{B})}{n}$ and $\frac{n(\text{B})}{n}$ denotes the ratio of the number of favourable outcomes to the total number of outcomes; therefore, it indicates the probability.

Therefore, $\frac{n(\text{A} \cap \text{B})}{n}$ can be written as $\text{P}(\text{A} \cap \text{B})$ and $\frac{n(\text{B})}{n}$ as $\text{P}(\text{B})$.

$=>\text{P}(\text{A} | \text{B}) = \frac{\text{P}(\text{A} \cap \text{B})}{\text{P}(\text{B})}$

Therefore, $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{B}) \text{P}(\text{A}| \text{B})$ if $\text{P}(\text{B}) \ne 0$

$= \text{P}(\text{A}) \text{P}(\text{B}| \text{A})$ if $\text{P}(\text{A}) \ne 0$

Similarly, the probability of occurrence of $\text{B}$ when $\text{A}$ has already occurred is given by $\text{P}(\text{B}| \text{A}) = \frac{\text{P}(\text{B} \cap \text{A})}{\text{P}(\text{A})}$.

Examples of Conditional Probability

Example 1: Two dice are rolled simultaneously and the sum of the numbers obtained is found to be $7$. What is the probability that the number $3$ has appeared at least once?

The sample space $\text{S}$ would consist of all the numbers possible by the combination of two dice. Therefore $\text{S}$ consists of $6 \times 6$ i.e. $36$ events.

Let event $\text{A}$ indicate the combination in which $3$ has appeared at least once.

And event $\text{B}$ indicates the combination of the numbers which sum up to $7$.

Therefore, $\text{A} = \{ (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3) \}$

And, $\text{B} = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \}$

Therefore, $\text{P}(\text{A}) = \frac{11}{36}$

And, $\text{P}(\text{B}) = \frac{6}{36}$

And also, $n(\text{A} \cap \text{B}) = 2$

Therefore, $\text{P}(\text{A} \cap \text{B}) = \frac{2}{36}$

Applying the conditional probability formula we get, $\text{P}(\text{A} | \text{B}) = \frac{\text{P}(\text{A} \cap \text{B})}{\text{P}(\text{B})} = \frac{\frac{2}{36}}{\frac{6}{36}} = \frac{2}{6} = \frac{1}{3}$.

Example 2: The probability that it is Friday and that a student is absent is $0.03$. Since there are $5$ school days in a week, the probability that it is Friday is $0.2$. What is the probability that a student is absent given that today is Friday?

In this problem, $\text{P}(\text{Friday and Absent}) = 0.03$

And, $\text{P}(\text{Friday}) = 0.2$

Therefore, the probability that a student is absent given that today is Friday $P(\text{Absent} | \text{Friday}) = \frac{\text{P}(\text{Friday and Absent})}{\text{P}(\text{Friday})} = \frac{0.03}{0.2} = \frac{3}{20} = 0.15$

Example 3: In a school, $18 \%$ of all students play football and basketball and $32 \%$ of all students play football. What is the probability that a student plays basketball given that the student plays football?

Let $\text{F}$ be an event that a student plays football.

And, let $\text{B}$ be an event that a student plays basketball.

Therefore, $\text{F} \cap \text{B}$ is an event that a student plays football as well as basketball.

$\text{P}(\text{F}) = 18\% = 0.18$

$\text{P}(\text{F} \cap \text{B}) = 32\% = 0.32$

Therefore, the probability that a student plays basketball given that the student plays football is $\frac{\text{P}(\text{F} \cap \text{B})}{\text{P}(\text{F})} = \frac{0.18}{0.32} = 0.5625$.

Rule 6: Multiplication Rule of Probability 

This probability rule is used when both events occur

$\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}| \text{B})$

To prove the multiplication rule of probability, we’ll use the following rules of conditional probability

$\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}) \times \text{P}(\text{B} | \text{A})$ ; if $\text{P}(\text{A}) \ne 0$

$\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{B}) \times \text{P}(\text{A}| \text{B})$ ; if $\text{P}(\text{B}) \ne 0$

If $\text{A}$ and $\text{B}$ are two independent events for a random experiment, then the probability of the simultaneous occurrence of two independent events will be equal to the product of their probabilities, i.e., $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}).\text{P}(\text{B})$

Now, we know that $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}) \times \text{P}(\text{B} | \text{A})$

Since $\text{A}$ and $\text{B}$ are independent, therefore, $\text{P}(\text{B} | \text{A}) = \text{P}(\text{B})$

Examples of Multiplication Rule of Probability

Example 1: An urn contains $12$ pink balls and 6 blue balls. Without replacement, two balls are drawn one after another. What is the probability that both balls drawn are pink?

Let $\text{A}$ and $\text{B}$ be the probability of drawing the first ball pink and the second ball also pink, respectively. Then, we need to find the probability of $\text{P} (\text{A} \cap \text{B})$.

The probability of drawing the first ball pink = $\frac{12}{18}$

Since the replacement is not allowed, and the first ball has been drawn, the remaining pink balls in the urn are $17$, out of which $11$ are pink balls.

Therefore, the probability of drawing another pink ball = $\frac{11}{17}$.

By multiplication rule, $\text{P} (\text{A} \cap \text{B}) = \text{P} (\text{A}) \times \text{P} (\text{A}| \text{B})$.

$= \frac{12}{18} \times \frac{11}{17} = \frac{132}{306}$.

Example 2: A magician takes out two cards from a deck of cards, one after the other, without replacement. What is the probability of getting an ace of spade, and a card of heart, as the first and second cards, respectively?

Since the cards are drawn without replacement, therefore, both events are dependent upon each other.

As we know, there is only one ace of spades in a deck of cards. Therefore, the probability of drawing ace of spades = $\frac{1}{52}$.

Since the ace of spades has already been drawn and replacement is not allowed. Therefore, only $51$ cards remain in the deck.

Out of these $51$ cards, $13$ are hearts. Therefore, the probability of drawing a card of heart = $\frac{13}{51}$.

From the multiplication rule, the probability of drawing aces of spades and a heart of card = $\frac{1}{52} \times \frac{13}{51} = \frac{1}{204}$. 

Practice Problems

1. In a cricket tournament, a player hits eight times ‘$6$’ out of thirty-two balls. Calculate the probability that he would not hit a ‘$6$’.
2. In a laptop shop, there are $16$ defective laptops out of $200$ laptops. If one laptop is taken out at random from this laptop shop, what is the probability that it is a non-defective laptop?
3. Neha has $4$ yellow t-shirts, $6$ black t-shirts, and $2$ blue t-shirts to choose from for her outfit today. She chooses a t-shirt randomly with each t-shirt equally likely to be chosen. Find the probability that a black or blue t-shirt is chosen for the outfit.
4. There are a total of $50$ distinct books on a shelf such as $20$ math books, $16$ physics books, and $14$ chemistry books. Find the probability of getting a book that is not a chemistry book or not a physics book.
5. Suppose you take out two cards from a standard pack of cards one after another, without replacing the first card. What is the probability that the first card is the ace of spades, and the second card is a heart?
6. You have a cowboy hat, a top hat, and an Indonesian hat called a songkok. You also have four shirts: white, black, green, and pink. If you choose one hat and one shirt at random, what is the probability that you choose the songkok and the black shirt?
7. Suppose a box has $3$ red marbles and $2$ black ones. We select $2$ marbles. What is the probability that the second marble is red given that the first one is red?
8. A family has $2$ children. Given that one of the children is a boy, what is the probability that the other child is also a boy?

FAQs

Conclusion

There are five basic probability rules that are used to solve problems. These five rules of probability are rules related to impossible and certain events, the sum of all probabilities, addition of probabilities, multiplication of probabilities, complementary events, and conditional probability.

Recommended Reading

• What is Experimental Probability – Definition, Approach, Formula & Examples
• What is Theoretical Probability – Definition, Approach, Formula & Examples
• What is Probability – Definition, Terminologies, Uses & Examples