Question

How does the General Multiplication rule differ from the Special Multiplication Rule of Probability?

Answer 1

Hint: In this problem we need to write the difference between the General Multiplication rule and Special Multiplication rule. For this we will discuss both the multiplication rules and write the formulas used in each rule. After that we will go for an example problem for both the rules.

Complete step by step solution:
General Multiplication Rule of Probability is something that is related to the probability of a combined occurrence of any two events let’s say $A$ and $B$ which is denoted by $A\centerdot B$. The expression for this rule depends on the probabilities of individual events and the conditional probabilities (probability of occurrence of one event under condition of occurrence of another). We can write it as Mathematically
$P\left( A\centerdot B \right)=P\left( A \right)\centerdot P\left( B|A \right)=P\left( B \right)\centerdot P\left( A|B \right)$
Special Multiplication Rule of Probability deals with the occurrence of two independent events. In this rule the result will be the product of the probabilities of each event. Mathematically we can write it as
$P\left( A\centerdot B \right)=P\left( A \right)\centerdot P\left( B \right)$
Example for Special Multiplication rule.
Let’s say the probability of the event $A$ is $P\left( A \right)=\dfrac{3}{9}$, probability of the event $B$ is $P\left( B \right)=\dfrac{2}{9}$, then the value of $P\left( A\centerdot B \right)$ or $P\left( A\cup B \right)$ will be
$\begin{align}
  & P\left( A\cup B \right)=P\left( A \right).P\left( B \right) \\
 & \Rightarrow P\left( A\cup B \right)=\dfrac{3}{9}\times \dfrac{2}{9} \\
 & \Rightarrow P\left( A\cup B \right)=\dfrac{2}{27} \\
\end{align}$
Example for General Multiplication rule.
Consider a bag containing $6$ black marbles and $4$ white marbles. Two marbles are drawn from the bag, without replacement. Now the probability that the both marbles are White will use the General Multiplication rule.
Now the probability of drawing a White ball from the bag is $P\left( A \right)=\dfrac{4}{10}$
Now the probability of drawing a White ball from the bag after removing a White ball which is drawn earlier is $P\left( B|A \right)=\dfrac{4-1}{10-1}=\dfrac{3}{9}$
Now the required probability is
$\begin{align}
  & P\left( A\centerdot B \right)=\dfrac{4}{10}\times \dfrac{3}{9} \\
 & \Rightarrow P\left( A\centerdot B \right)=\dfrac{2}{15} \\
\end{align}$

Note:
Whenever you go with this problem the key term you should mention is Independent events. A Special Multiplication Rule is used for the Independent event where the General Multiplication Rule is used for any event.