Definition of Probability :

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$\quad$ If an experiment results in a total of $(m+n)$ outcomes which are equally likely and mutually exclusive with one another and

$\quad$ if ’ $m$ ’ outcomes are favorable to an event ’ $A$ ’ while ’ $n$ ’ are unfavorable, then

$\quad$ the probability of occurrence of the event ‘$A$’ $${P(A)=\frac{m}{m+n}=\frac{n(A)}{n(S)}}$$

$\quad$ We say that odds in favour of ’ $A$ ’ are $m: n$, while odds against ’ $A$ ’ are $n: m$.

$${P(\bar{A})=\frac{n}{m+n}=1-P(A)}$$

Addition theorem of probability :

$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$

De Morgan’s Laws:

• $ (A \cup B)^{c}=A^{c} \cap B^{c}$

• $ (A \cap B)^{c}=A^{c} \cup B^{c}$

Distributive Laws:

PYQ-2023-Quadratic_Equation-Q10, PYQ-2023-Probability-Q14

• $ A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$

• $ A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$

Important Formula:

• $ P(A$ or $B$ or $C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A)+P(A \cap B \cap C)$

• $ P$ (at least two of $A, B, C$ occur $)=P(B \cap C)+P(C \cap A)+P(A \cap B)-2 P(A \cap B \cap C)$

• $ P($ exactly two of $A, B, C$ occur $)=P(B \cap C)+P(C \cap A)+P(A \cap B)-3 P(A \cap B \cap C)$

• $ P($ exactly one of $A, B, C$ occur $)=P(A)+P(B)+P(C)-2 P(B \cap C)-2 P(C \cap A)-2 P(A \cap B)+3 P(A \cap B \cap C)$

Conditional probability:

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• $P(A | B)=$ Probability of occurrence of $A$, given that $B$ has already happened = $$\frac{P(A \cap B)}{P(B)}$$

• $P(B | A)=$ Probability of occurrence of $B$, given that $A$ has already happened = $$\frac{P(A \cap B)}{P(A)}$$

$\quad$ Note:

• If the outcomes of the experiment are equally likely, then $$ P(A | B)=\frac{\text { Number of sample points in } (A \cap B)}{\text { Number of points in } B} $$

• If $A$ and $B$ are independent events, then $P(A | B)=P(A)$ and $P(B | A)=P(B)$

Binomial distribution:

PYQ-2023-Probability-Q13

$ \quad$ If an experiment is repeated $n$ times, the successive trials being independent of one another, then

$ \quad$ the probability of $r$ success is ${ }^n C_r p^r q^{n-r}$

$ \quad$ the probability of at least $r$ success is $\sum_{k=r}^n{ }^n C_k p^k q^{n-k},$

$ \quad$ where $p$ is probability of success in a single trial, $q=1-p$

$\quad \quad \quad$ • Mean $\mathrm{E}(\mathrm{x})=\mathrm{np}$

$\quad \quad \quad$ • $\mathrm{E}(\mathrm{x}^2)=\mathrm{npq}+\mathrm{n}^2 \mathrm{p}^2$

$\quad \quad \quad$ • Variance $\mathrm{E}(\mathrm{x}^ 2)-(\mathrm{E}(\mathrm{x}))^2=\mathrm{npq}$

$\quad \quad \quad$ • Standard deviation $=\sqrt{\mathrm{npq}}$

Expectation :

$\quad$ If a value $M_{i}$ is associated with a probability of $p_{i}$, then the expectation is given by $\Sigma p_{i} M_{i}$

Total Probability Theorem :

$$\quad P(A)=\sum_{i=1}^{n} P\left(B_{i}\right) \cdot P\left(A | B_{i}\right)$$

Bayes’ Theorem :

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$\quad$ If an event $A$ can occur with one of the $n$ mutually exclusive and exhaustive events $B_{1}, B_{2}, \ldots \ldots, B_{n}$ and

$\quad$ the probabilities $P\left(A | B_{1}\right), P\left(A | B_{2}\right) \ldots . P\left(A | B_{n}\right)$ are known, then

$$P\left(B_{i} | A\right)=\frac{P\left(B_{i}\right) \cdot P\left(A | B_{i}\right)}{\sum_{i=1}^{n} P\left(B_{i}\right) \cdot P\left(A | B_{i}\right)}$$

• $A=\left(A \cap B_{1}\right) \cup\left(A \cap B_{2}\right) \cup\left(A \cap B_{3}\right) \cup \ldots \ldots . \cup\left(A \cap B_{n}\right)$

• $P(A)=P\left(A \cap B_{1}\right)+P\left(A \cap B_{2}\right)+\ldots \ldots+P\left(A \cap B_{n}\right)=\sum_{i=1}^{n} P\left(A \cap B_{i}\right)$

Binomial Probability Distribution :

• Mean of any probability distribution of a random variable is given by $$\mu=\frac{\Sigma p_{i} x_{i}}{\Sigma p_{i}}=\Sigma p_{i} x_{i}$$

• Variance of a random variable is given by $$\sigma^{2}=\Sigma\left(x_{i}-\mu\right)^{2} \cdot p_{i}=\Sigma p_{i} x_{i}^{2}-\mu^{2}$$

Mathematical definition of probability:

$$ \text { Probability of an event }=\frac{\text { Number of favorable cases to event A }}{\text { Total Number of cases }} $$

• Probability with replacement: 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.

• Probability without replacement: When sampling is done without replacement

Probability of a union:

$$\mathbb{P}(A \cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A \cap B)$$

$\quad \quad$ For three events $A, B, C$ : $$ \mathbb{P}(A \cup B \cup C)=\mathbb{P}(A)+\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(A \cap B)-\mathbb{P}(A \cap C)-\mathbb{P}(B \cap C)+\mathbb{P}(A \cap B \cap C) $$

Note:

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• $0 \leq P(A) \leq 1$

• Probability of an impossible event is zero

• Probability of a sure event is one.

• $P(A)+P($ Not $A)=1$ i.e. $P(A)+P(\bar{A})=1$

Odd for an event:

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$\quad$ If $P(A)=\frac{m}{n}$ and $P(\bar{A})=\frac{n-m}{n}$ Then,

$\quad \quad \quad$ • odds in favor of $A=\frac{P(A)} {P(\bar{A})}=\frac{m}{n-m}$ and

$\quad \quad \quad$ • odd in against of $A=\frac{p(\bar{A})}{P(A)}=\frac{n-m}{m}$ placement, each member of a population may be chosen only once.

Set theoretical notation of probability and some important results:

PYQ-2023-Probability-Q8, PYQ-2023-Probability-Q13

• $P(A \cup B)=1-P(\bar{A} \cap \bar{B})$

• $P(A \ B)=\frac{P(A \cap B)}{P(B)}$

• $P(A \cup B)=P(A \cap B)+P(\bar{A} \cap B)+P(A \cap \bar{B})$

• $A \subseteq B \Rightarrow P(A) \subseteq P(B)$

• $P(\bar{A} \cap B)=P(B)-P(A \cap B)$

• ${P}({A} \cap {B}) \leq {P}({A}) {P}({B}) \leq P(\mathrm{A} \cup \mathrm{B}) \leq \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})$

• $P($ Exactly one event $)=P(A \cap \bar{B})+P(\bar{A} \cap B)$

• $P(\bar{A} \cup \bar{B})=1-P(A \cap B)=P(A)+P(B)-2 P(A \cap B)=P(A+B)-P(A \cap B)$

• $\mathrm{P}($ neither $\mathrm{A}$ nor $\mathrm{B})=\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$

• $\mathrm{P}($ none $)=1-\mathrm{P}$ (at least one)

Playing cards:

• Total cards: 52 (26 red, 26 black)

• Four suits: Heart, diamond, spade, club (13 cards each)

• Court (face) cards: 12 (4 kings, 4 queens, 4 jacks)

• Honor cards: 16 (4 Aces, 4 kings, 4 queens, 4 Jacks)

• Probability regarding cards:

  • When $n$ cards are drawn $(1 \leq n \leq 52)$ from well shuffled deck of 52 cards, the probability of each simple event is $\frac{1}{{ }^{52} C_n}$.

  • If $\mathrm{n}$ cards are drawn one after the other with replacement, the probability of each simple event is $\frac{1}{(52)^n}$

Probability with dice:

$\quad$ When a dice is rolled $n$ times or $n$ dice are rolled once, the probability of each simple event is $\frac{1}{6^n}$

Coin Toss Probability:

$\quad$ When a coin is tossed $n$ times or $n$ coins are tossed once, the probability of each simple event is $\frac{1}{2^n}$

Probability regarding $n$ letters and their envelopes:

$\quad$ If $n$ letters corresponding to $n$ envelopes are placed in the envelopes at random, then

• Probability that all letters are in the right envelopes $=\frac{1}{\mathrm{n} !}$

• Probability that all letters are not in the right envelopes $=1-\frac{1}{n !}$

• Probability that no letter is in the right envelope $$=\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots . .+(-1)^n \frac{1}{n !}$$

• Probability that $r$ letters are in the right envelope $$=\frac{1}{r !}\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots+(-1)^{n-r} \frac{1}{(n-r) !}\right]$$

Addition Theorem of Probability:

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• When events are mutually exclusive $$ \begin{aligned} & \text { i.e. } n(A \cap B)=0 \quad \Rightarrow P(A \cap B)=0 \ & \therefore P(A \cup B)=P(A)+P(B) \end{aligned} $$

• When events are not mutually exclusive i.e. $P(A \cap B) \neq 0$ $$ \therefore P(A \cup B)=P(A)+P(B)-P(A \cap B) \text { or }$$

$$P(A+B)=P(A)+P(B)-P(A B) $$

• When events are independent i.e. $P(A \cap B)=P(A) P(B)$ $$ \therefore P(A \cup B)=P(A)+P(B)-P(A) P(B) $$

Multiplication Theorem:

$$ P(A \cap B)=P(A | B) . P(B); \quad P(B) \neq 0 \text { or }$$

$$ P(A \cap B)=P(B | A) P(A); \quad P(A) \neq 0 $$

$\quad$ Generalized:

$$P\left(E_1 \cap E_2 \cap E_3 \cap \ldots \cap E_n\right) =P\left(E_1\right) P\left(E_2 | E_1\right) P\left(E_3 | E_1 \cap E_2\right) P\left(E_4 | E_1 \cap E_2 \cap E_3\right)… $$

• If events are independent, then $$ P\left(E_1 \cap E_2 \cap E_3 \ldots \cap E_n\right)=P\left(E_1\right) P\left(E_2\right) \ldots P\left(E_n\right) $$

Probability of at least one of the $\mathbf{n}$ Independent events:

PYQ-2023-Probability-Q14

$\quad$ If $\mathrm{P} _{1}, \mathrm{P} _{2}, \ldots ,\mathrm{P} _{\mathrm{n}}$ are the probabilities of $\mathrm{n}$ independent events $A _1, A _2 \ldots A _n$ then the probability that at least one of these events will happen is $$1-\left[\left(1-P _1\right)\left(1-P _2\right) \ldots\right. \left.\left(1-P _n\right)\right]$$

or $$P\left(A _1+A _2+\ldots+A _n\right)=1-P\left(\bar{A} _1\right) P\left(\bar{A} _2\right) \ldots P\left(\bar{A} _n\right)$$

Total probability:

$\quad$ Let $A_1, A_2 \ldots A_n$ be $n$ mutually exclusive & set of exhaustive events. If event $A$ can occur through any one of these events, then the probability of occurrence of $A$ $$ P(A)=P\left(A \cap A_1\right)+P\left(A \cap A_2\right)+\ldots+P\left(A \cap A_n\right)=\sum_{r=1}^n P\left(A_r\right) P\left(A / A_r\right) $$

Bayes’ Rule:

$\quad$ Let $A_{1^{\prime}} A_{2^{\prime}} A_3$ be any three mutually exclusive & exhaustive events (i.e. $A_1 \cup A_2 \cup A_3=$ sample space and $ A_1 \cap A_2 \cap A_3=\phi$ ) of a sample space $S$ and

$\quad$ $B$ is any other event on sample space then, $$ P\left(A_i | B\right)=\frac{P\left(B | A_i\right)\left(P\left(A_i\right)\right.}{P\left(B | A_1\right) P\left(A_1\right)+P\left(B | A_2\right) P\left(A_2\right)+P\left(B | A_3\right) P\left(A_3\right)}, i=1,2,3 $$

Probability distribution:

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$\quad$ If a random variable $x$ assumes values $x_{1}, x_{2}, \ldots x_n$ with probabilities $P_{1}, P_{2}, \ldots ,P_n$ respectively then

$\quad \quad \quad$ • $P_1+P_2+P_3+\ldots+P_n=1$

$\quad \quad \quad$ • Mean = $\mathrm{E}(\mathrm{x})=\Sigma \mathrm{P} _{\mathrm{i}} \mathrm{x} _{\mathrm{i}}$

$\quad \quad \quad$ • Variance $=\sum x^2 P_i-(\text { mean })^2=\sum\left(x^2\right)-(E(x))^2$

Truth of the statement:

• If two persons $A$ and $B$ speak the truth with probabilities $P_1$ and $ P_2$ respectively and if they agree on a statement, then the probability that they are speaking the truth will be given by $$\frac{P_1 P_2}{P_1 P_2+\left(1-P_1\right)\left(1-P_2\right)}$$.

• If A and B both assert that an event has occurred, the probability of occurrence of which is $\alpha$, then the probability that the event has occurred is $$\frac{\alpha P_1 P_2}{\alpha P_1 P_2+(1-\alpha)\left(1-P_1\right)\left(1-P_2\right)}$$

  given that the probability of $A $ and $ B$ speaking truth is $\mathrm{P}_1, \mathrm{P}_2$ respectively.

• If in the second part, the probability that their lies coincide is $\beta$, then from the above case, the required probability will be $$\frac{\alpha P_1 P_2}{\alpha P_1 P_2+(1-\alpha)\left(1-P_1\right)\left(1-P_2\right) \beta}$$