EASYAmity University, Mumbai B.Tech Admissions 2025
ApplyRanked amongst top 3% universities globally (QS Rankings)
Multiplication Theorem on Probability - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
-
26 Questions around this concept.
Solve by difficulty
A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same color are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is:
A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is
What is the probability of getting one facecard from a deck and 2 in a dice throw if these two are dependent events such that probability of obtaining a face card is 1/12 if 2 occurs in a dice throw ?
Concepts Covered - 1
Multiplication Theorem on Probability
Let $A$ and $B$ are two events associated with a sample space $S$. The set $A \cap B$ denotes the event that both $A$ and $B$ have occurred. In other words, $A \cap B$ denotes the simultaneous occurrence of the events $A$ and $B$. The event $A \cap B$ is also written as $A B$.
The probability of event $A B$ or $A \cap B$ can be obtained by using the conditional probability.
From the conditional probability of event A given that B has occurred is denoted by $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$ and is given by
$
\mathrm{P}(\mathrm{~A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}, \mathrm{P}(\mathrm{~B}) \neq 0
$
Using this result, we can write
$
\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\mathrm{P}(\mathrm{~B}) \cdot \mathrm{P}(\mathrm{~A} \mid \mathrm{B})
$
Also, we know that
$
\begin{aligned}
\mathrm{P}(\mathrm{~B} \mid \mathrm{A}) & =\frac{\mathrm{P}(\mathrm{~B} \cap \mathrm{~A})}{\mathrm{P}(\mathrm{~A})}, \mathrm{P}(\mathrm{~A}) \neq 0 \\
\text { or } \quad \mathrm{P}(\mathrm{~B} \mid \mathrm{A}) & =\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}, \mathrm{P}(\mathrm{~A}) \neq 0 \quad(\because \mathrm{~A} \cap \mathrm{~B}=\mathrm{B} \cap \mathrm{~A})
\end{aligned}
$
Thus, $\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B} \mid \mathrm{A})$
Combining (1) and (2), we get
$
\begin{aligned}
\mathrm{P}(\mathrm{~A} \cap \mathrm{~B}) & =\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}(\mathrm{~B} \mid \mathrm{A}) \\
& =\mathrm{P}(\mathrm{~B}) \cdot \mathrm{P}(\mathrm{~A} \mid \mathrm{B}) \quad(\text { provided } \mathrm{P}(\mathrm{~A}) \neq 0 \text { and } \mathrm{P}(\mathrm{~B}) \neq 0)
\end{aligned}
$
The above result is known as the multiplication rule of probability.
Multiplication rule of probability for more than two events
If $A, B$ and $C$ are three events associated with sample space, then we have
$
\begin{aligned}
P(A \cap B \cap C) & =P(A) P(B \mid A) P(C \mid A \cap B) \\
& =P(A) P(B \mid A) P(C \mid A B)
\end{aligned}
$
Similarly, the multiplication rule of probability can be extended for four or more events.
"Stay in the loop. Receive exam news, study resources, and expert advice!"
Get Answer to all your questions
JEE Main Articles
Back to top