When we come across events that are not independent, ‘conditional’ events are plausible and account for the possible change in the multiplication of probability. Recall that if A and B are independent, then we can accept that :
P(A ∩ B) = P(A) * P(B)
However, if P(B) changes based on the occurrence of event A, then these events are not independent.There is a special notation to explain this dependence: P(B | A). Hence, we use conditional probability to obtain the general multiplication rule.
P(A ∩ B) = P(A) * P(B|A)
This version of the multiplication rule is not limited to independent events.
Example: What is the probability of randomly drawing a card from the deck that is a king AND a Heart?
P(king and heart) = P(king) * P(heart | king)
P(king) = (4/52)
P(Heart | king) = (1/4)
Answer: = (4/52) * (1/4) = 1/52
Example: A box n contains 15 black and 10 white balls. Two balls are drawn from the box one after the other and they are not replaced between events. What is the probability that both drawn balls are white?
Solution Let P and Q denote that a white ball is drawn from the box from the first and second draw respectively without replacement.
Then P (P) = P ( white ball in the first draw) = 10/25
Now that event P has occurred
P (Q) is conditional therefore: P (Q│P) = 9/24
Now we can apply the multiplication rule, such that
P (P ∩ Q) = P (P) ∙ P(Q│P) = $\frac{10}{25}\cdot
\frac{9}{24}=\frac{3}{20}$
When more than two events: A, B, and C are involved in a sample space: S, then:
P(A ∩B∩C) = P(A) P(B|A) P(C|(A∩B)) = P(A) P(B|A) P(C|AB)
Example: Three cards are drawn successively, without replacement from a deck of 52 unbiased cards. What is the probability that the first two cards are Queens and the third card drawn is a king?
Solution: Let Q denote the event that the card drawn is a queen and K be the event that the card drawn is a king. From the question, we have to find P (QQK)
We know that P(Q) = 4/52 for the first draw
For second draw P (Q│Q) = 3/51
For third draw P (K│QQ) = 4/50
Now we can use the multiplication law:
P (QQK) = P (Q) ∙ P (Q│Q) ∙ P (K│QQ) = $\frac{4}{52}\cdot \frac{3}{51}\cdot \frac{4}{50}=\frac{2}{5525}$