Sample Space (S): Set of all possible outcomes.
Event (E): Any subset of sample space.
Favourable Outcomes: Outcomes belonging to the event.
Probability of an event:
$$P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}$$
Find the probability of getting a head in a coin toss: $$P(H)=\frac{1}{2}$$
All outcomes have equal chance.
If two events cannot occur at the same time:
$$P(A \cap B) = 0$$
Events covering whole sample space.
Occurrence of one event does not affect the other. $$P(A \cap B)=P(A)\,P(B)$$
For any events A and B:
$$P(A \cup B)=P(A)+P(B)-P(A\cap B)$$
If events do not occur together: $$P(A \cup B)=P(A)+P(B)$$
For any events: $$P(A\cap B)=P(A)\cdot P(B|A)$$
If A and B are independent: $$P(A\cap B)=P(A)\,P(B)$$
Probability of A given that B has happened: $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
Two cards drawn from a deck. If first is Ace, find probability second is also Ace: $$P(A_2|A_1)=\frac{3}{51}$$
Used to reverse conditional probabilities. $$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\bar{A})P(\bar{A})}$$
A variable representing numerical value of an outcome. Types: • Discrete • Continuous
For discrete X: $$P(X=x) = p(x)$$ and $$\sum p(x)=1$$
$$E(X)=\sum x\,p(x)$$
$$Var(X)=E(X^2)-E(X)^2$$
PMF: $$P(X=1)=p, \; P(X=0)=1-p$$ Mean: $p$ Variance: $p(1-p)$
$$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$ Mean: $np$ Variance: $np(1-p)$
$$P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}$$ Mean = Variance = $\lambda$
$$f(x)=\frac{1}{b-a},\; a\le x\le b$$ Mean: $\frac{a+b}{2}$ Variance: $\frac{(b-a)^2}{12}$
1. $0 \le P(E) \le 1$
2. $P(S)=1$, $P(\phi)=0$
3. $P(\bar{E})=1-P(E)$
4. $P(A\cap B)=P(A)+P(B)-P(A\cup B)$
5. For independent events: $P(A|B)=P(A)$
Q1. A die is rolled. Find probability that outcome is a multiple of 3.
Solution: Fav = {3,6}, Total = 6 → $$P=\frac{2}{6}=\frac{1}{3}$$
Q2. Two coins tossed. Probability of at least one head?
Total = 4, favourable = 3 → $$P=\frac{3}{4}$$
Q3. A bag has 3 red and 5 blue balls. 2 drawn. Probability both red?
$$P=\frac{\binom{3}{2}}{\binom{8}{2}}=\frac{3}{28}$$
Q4. If $P(A)=0.4$, $P(B)=0.5$, $P(A\cap B)=0.2$ find $P(A\cup B)$.
$$P=0.4+0.5-0.2=0.7$$
Q5. In a binomial distribution $n=5$, $p=0.4$, find $P(X=2)$.
$$P=\binom52(0.4)^2(0.6)^3=0.3456$$
These probability inequalities are useful for competitive exams like NIMCET and JEE Mains. You can directly use this content in your nimcet notes and jee mains notes.
For any event $A$: $$0 \le P(A) \le 1$$
For an event $A$ and its complement $\bar{A}$: $$P(A) + P(\bar{A}) = 1$$ Hence, $$P(\bar{A}) = 1 - P(A)$$ and $$0 \le P(\bar{A}) \le 1$$
For any two events $A$ and $B$:
Union always increases possibilities: $$P(A \cup B) \ge P(A), \quad P(A \cup B) \ge P(B)$$
Intersection always reduces possibilities: $$P(A \cap B) \le P(A), \quad P(A \cap B) \le P(B)$$
Using the inclusion–exclusion principle: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Since $P(A \cap B) \ge 0$: $$P(A \cup B) \le P(A) + P(B)$$
Also, as $P(A \cup B) \ge 0$ and $P(A \cap B) \le 1$: $$P(A) + P(B) - 1 \le P(A \cup B) \le P(A) + P(B)$$
For events $A_1, A_2, \dots, A_n$: $$P(A_1 \cup A_2 \cup \dots \cup A_n) \le \sum_{i=1}^{n} P(A_i)$$ This is a general form of the union bound.
For any number of events $A_1, A_2, A_3, \dots$: $$P\left(\bigcup_{i=1}^{\infty} A_i\right) \le \sum_{i=1}^{\infty} P(A_i)$$ This is very useful when the intersections are not known.
For two events $A$ and $B$: $$P(A) + P(B) - 1 \le P(A \cap B)$$ because $P(A \cup B) \le 1$ and $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
If $X$ is a non-negative random variable and $a > 0$, then: $$P(X \ge a) \le \frac{E(X)}{a}$$ This gives an upper bound on the tail probability using only the expectation.
If $X$ is a random variable with mean $\mu$ and variance $\sigma^2$, then for any $k > 0$: $$P(|X - \mu| \ge k\sigma) \le \frac{1}{k^2}$$ This inequality shows that most of the probability mass lies within a few standard deviations of the mean.
Let $X$ be a random variable and $\phi$ be a convex function. Then: $$\phi(E(X)) \le E(\phi(X))$$ This inequality is widely used in advanced probability and statistics, information theory, and optimization.
Disclaimer : This Information is Only Immediate Information. Complete Caution has been Taken to Create This Notification, But if There are Any Errors, We will not be Responsible. Therefore, Before filling the Form, Please Visit the Relevant Official Website and Read the Entire Information Carefully. We are Looking Forward to Updating From Time to Time but Still You Should Match the Official Website. If There is a Change in Date or Information, Contact us. In Case of Any Loss, the Student can not Submit the Information Given on This Website's Page as a Legal Document.