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Independent Event is considered one of the most asked concept.
64 Questions around this concept.
Let A and B two events such that $P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{1}{4}$ where $\bar{A}$ stands for the complement of the event $A$. Then the events $A$ and $B$ are :
Let $A$ and $B$ be independent events such that $P(A)=p, P(B)=2 p$. The largest value of $p$, for which P (exactly one of $\mathrm{A}, \mathrm{B}$ occurs) $=\frac{5}{9}$, is :
Consider the following statements
P :Suman is brilliant
Q: Suman is rich
R: Suman is honest.
The negative of the statement."Suman is brilliant and dishonest, if and only if Suman is rich" can be expressed as
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Let Ajay will not appear in JEE exam with probability $\mathrm{p}=\frac{2}{7}$, while both Ajay and Vijay will appear in the exam with probability $\mathrm{q}=\frac{1}{5}$. Then the probability, that Ajay will appear in the exam and Vijay will not appear is :
If $A$ and $B$ are independent events then $\bar{A}$ and $\bar{B}$ are
For throwing a dice, which is a simple event?
For throwing a dice which is a compound event?
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In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is
Let $\mathrm{S}=\{1,2,3, \ldots, 2022\}$. Then the probability, that a randomly chosen number n from the set S such that $\operatorname{HCF}(\mathrm{n}, 2022)=1$, is :
If the events A and B are independent, then P(A ∩ B) is equal to
Two or more events are said to be independent if the occurrence or non-occurrence of any of them does not affect the probability of occurrence or non-occurrence of other events.
Two events $A$ and $B$ are said to be independent, if
1. $P(A \mid B)=P(A)$
2. $P(B \mid A)=P(B)$
A third result can also be obtained for independent events
From the multiplication rule of probability, we have
$
P(A \cap B)=P(A) P(B \mid A)
$
Now if $A$ and $B$ are independent, then $P(B \mid A)=P(B)$, so
$
\text { 3. } P(A \cap B)=P(A) \cdot P(B)
$
To show two events are independent, you must show only one of the above three conditions.
If two events are NOT independent, then we say that they are dependent.
With and Without replacement
In some questions, like the ones related to picking some object from a bag with different kinds of objects in it, objects may be picked with replacement or without replacement.
With replacement: If each object re-placed in the box after it is picked, then that object has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
Without replacement: When sampling is done without replacement, then probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.
Difference between Independent events and Mutually Exclusive events
$A$ and $B$ are mutually exclusive events if they cannot occur at the same time. So, if $A$ occurred then $B$ cannot occur and vice versa. This means that $A$ and $B$ do not share any outcomes and $P(A \cap B)=0$. Also, $P(A / B)=0$ and $P(B / A)=0$.
But in case of independent events $A$ and $B, P(A / B)=P(A)$ [and not 0 as in case of mutually exclusive events], and $P(B / A)=P(B)$ [not 0]
Three Independent Events
Three events A, B and C are said to be mutually independent, if
$
\begin{aligned}
& P(A \cap B)=P(A) \cdot P(B) \\
& P(A \cap C)=P(A) \cdot P(C) \\
& P(B \cap C)=P(B) \cdot P(C) \\
& P(A \cap B \cap C)=P(A) \cdot P(B) \cdot P(C)
\end{aligned}
$
and
If at least one of the above is not true for three given events, we say that the events are not independent.
Properties of Independent Event
If $A$ and $B$ are independent events, then
$
\text { 1. } \begin{aligned}
P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\
& =P(A)+P(B)-P(A) \cdot P(B)
\end{aligned}
$
2. Event $A^{\prime}$ and $B$ are independent.
$
\begin{aligned}
P\left(A^{\prime} \cap B\right) & =P(B)-P(A \cap B) \\
& =P(B)-P(A) P(B) \\
& =P(B)(1-P(A)) \\
& =P(B) P\left(A^{\prime}\right)
\end{aligned}
$
3. Event A and $\mathrm{B}^{\prime}$ are independent.
4. Event $A^{\prime}$ and $B^{\prime}$ are independent.
$
\begin{aligned}
P\left(A^{\prime} \cap B^{\prime}\right) & =P\left((A \cup B)^{\prime}\right) \\
& =1-P(A \cup B) \\
& =1-P(A)-P(B)+P(A) \cdot P(B) \\
& =(1-P(A))(1-P(B)) \\
& =P\left(A^{\prime}\right) P\left(B^{\prime}\right)
\end{aligned}
$
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