12.3 Multiplication Law of Probability

12.3 Multiplication Law of Probability

(i) If $A$ and $B$ are independent events, then

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

(ii) Conditional Probability

If $A$ and $B$ are dependent events, then

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

where $P(B | A)$ represents the probability of event $B$ occurring after it is assumed that event $A$ has already occurred (read $B | A$ as "$B$ given $A$").

Example 11

A bag contains $4$ red and $3$ green balls. Kabeer draw two balls one after the other with replacement (the second is drawn after the first is replaced). Find the probability that the first ball is red and the second ball is also red.

Solution Let $A$ be the event that first ball is red and $B$ be the event that second ball is also red.

Then $P(A) = \dfrac{4}{7}$ and $P(B) = \dfrac{4}{7}$ (Since the ball is replaced, the sample space is not affected).

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

$= \dfrac{4}{7} \times \dfrac{4}{7} = \dfrac{16}{49}$

Example 12

A bag contains $3$ white and $2$ black marbles. Two marbles are drawn one after the other without replacement. Find the probability that the first marble is white and the second marble is also white.

Solution Let $A$ be the event that first marble is white and $B$ be the event that second marble is also white.

Total marbles $= 3 + 2 = 5$

Then $P(A) = \dfrac{3}{5}$ and $P(B | A) = \dfrac{2}{4}$ (Since the marble is not replaced, the sample space is affected).

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

$= \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$

Example 13

Majid dealt two cards successively (without replacement) from a well shuffled deck of $52$ playing cards. Find the probability that the first card is a king and the second card is a queen.

Solution Let $A$ be the event that drawing card is a king and $B$ be the event that drawing card is a queen. Remember that we must take care for the sample space of event $B$, that event $A$ has already occurred.

$P(A) = \dfrac{4}{52} = \dfrac{1}{13} ;\ P(B | A) = \dfrac{4}{51}$

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

$= \dfrac{1}{13} \times \dfrac{4}{51} = \dfrac{4}{663}$