Question

How does the general multiplication rule differ from the special multiplication rule of probability?

Answer 1

Step-1: General multiplication rule of probability:

For any two events $A$ and $B$ ($A\&B$ are dependent).

$P\left(A\cap B\right)=P\left(A\right)\times P\left(\frac{B}{A}\right)$, $...\left[1\right]$

Where $P\left(\frac{B}{A}\right)=$ Conditional probability of $B$, taking into account that $A$ has occurred.

For example -

Choose two cards randomly from a deck of cards without replacing the first card before the second card is picked. What is the probability of picking a spade in both attempts?

Let $E_1$ be the event to pick a spade in the first attempt.

And $E_2$ be the event to pick a spade in the second attempt.

Sampling is done without replacement.

$\therefore$ $E_2$ depends on $E_1$.

$\Rightarrow$ $P\left(E_1\right)=\frac{13}{52}$

$\Rightarrow$ $P\left(\frac{E_2}{E_1}\right)=\frac{12}{51}$

Apply multiplication rule for dependent events

$P\left(E_1\cap E_2\right)=P\left(E_1\right)\times P\left(\frac{E_2}{E_1}\right)$

$=\frac{13}{52}\times \frac{12}{51}$

$=\frac{1}{17}$

Hence, $\frac{1}{17}$ of the selection will have a spade in first and then a spade in the second attempt.

Step-2: Specific multiplication rule of probability:

For any two events $A$ and $B$ ($A\&B$ are independent)

Since the occurrence of $A$ would have no effect on the probability of $B$ occurring.

$\therefore$ $P\left(\frac{B}{A}\right)=P\left(B\right)$

Equation $\left[1\right]$ can be written as

$P\left(A\cap B\right)=P\left(A\right)\times P\left(B\right)$

For example -

Choose two cards randomly from a deck of cards by replacing the first card before the second card is picked. What is the probability of picking a spade in both attempts?

Let $E_1$ be the event to pick a spade on the first selection.

And $E_2$ be the even to pick a spade on the second selection.

Sampling is done with replacement

$\therefore$ $E_2$ does not depend on $E_1$.

$P\left(E_1\right)=\frac{13}{52}$

$P\left(E_2\right)=\frac{13}{52}$

Apply specific multiplication rule for independent events.

$P\left(E_1\cap E_2\right)=P\left(E_1\right)\times P\left(E_2\right)$

$=\frac{13}{52}\times \frac{13}{52}$

$=\frac{1}{16}$

Hence, $\frac{1}{16}$ of the selection will have a spade in the first and then a spade in the second attempt.

Hence, this is how general and specific multiplication rules of probability differ.