Register

Web Page

Syllabus: Random variables, Uniform, Normal, Exponential, Poisson and Binomial distributions. Mean, median, mode and standard deviation. Conditional probability and Bayes theorem

$$\scriptsize{\overset{{\large{\textbf{Mark Distribution in Previous GATE}}}}{\begin{array}{|c|c|c|c|c|c|c|c|}\hline
\textbf{Year}& \textbf{2022} & \textbf{2021-1}&\textbf{2021-2}&\textbf{2020}&\textbf{2019}&\textbf{2018}&\textbf{2017-1}&\textbf{2017-2}&\textbf{2016-1}&\textbf{2016-2}&\textbf{Minimum}&\textbf{Average}&\textbf{Maximum}
\\\hline\textbf{1 Mark Count} & 0 &1&1&0&2&1&1&0&1&1&0&0.8&2
\\\hline\textbf{2 Marks Count} & 0 &2&2&1&1&1&0&3&1&0&0&1.1&3
\\\hline\textbf{Total Marks} & 0 &5&5&2&4&3&1&6&3&1&\bf{0}&\bf{3}&\bf{6}\\\hline
\end{array}}}$$

Filter

Highest voted questions in Probability

16 votes
3 answers
92
GATE CSE 2004 | Question: 25
If a fair coin is tossed four times. What is the probability that two heads and two tails will result? $\frac{3}{8}$ $\frac{1}{2}$ $\frac{5}{8}$ $\frac{3}{4}$
If a fair coin is tossed four times. What is the probability that two heads and two tails will result?$\frac{3}{8}$$\frac{1}{2}$$\frac{5}{8}$$\frac{3}{4}$
User Avatar
14 votes
1 answer
99
TIFR CSE 2011 | Part A | Question: 7
Let $X$ and $Y$ be two independent and identically distributed random variables. Then $P\left ( X> Y \right )$ is. $\frac{1}{2}$ 1 0 $\frac{1}{3}$ Information is insufficient.
Let $X$ and $Y$ be two independent and identically distributed random variables. Then $P\left ( X Y \right )$ is.$\frac{1}{2}$10$\frac{1}{3}$Information is insufficient.
User Avatar
14 votes
4 answers
100
TIFR CSE 2011 | Part A | Question: 3
The probability of three consecutive heads in four tosses of a fair coin is $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{8}\right)$ $\left(\dfrac{1}{16}\right)$ $\left(\dfrac{3}{16}\right)$ None of the above
The probability of three consecutive heads in four tosses of a fair coin is$\left(\dfrac{1}{4}\right)$$\left(\dfrac{1}{8}\right)$$\left(\dfrac{1}{16}\right)$$\left(\dfrac...
User Avatar
14 votes
2 answers
101
TIFR CSE 2010 | Part A | Question: 6
Given 10 tosses of a coin with probability of head = .$4$ = ($1$ - the probability of tail), the probability of at least one head is? $(.4)^{10}$ $1 - (.4)^{10}$ $1 - (.6)^{10}$ $(.6)^{10}$ $10(.4) (.6)^{9}$
Given 10 tosses of a coin with probability of head = .$4$ = ($1$ - the probability of tail), the probability of at least one head is?$(.4)^{10}$$1 - (.4)^{10}$$1 - (.6)^{...
User Avatar
14 votes
5 answers
102
GATE CSE 1995 | Question: 1.18
The probability that a number selected at random between $100$ and $999$ (both inclusive) will not contain the digit $7$ is: $\dfrac{16}{25}$ $\left(\dfrac{9}{10}\right)^{3}$ $\dfrac{27}{75}$ $\dfrac{18}{25}$
The probability that a number selected at random between $100$ and $999$ (both inclusive) will not contain the digit $7$ is: $\dfrac{16}{25}$$\left(\dfrac{9}{10}\right)^...
User Avatar
13 votes
2 answers
103
GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 4
$A$ and $B$ are two events. If $P(A, B)$ decreases while $P(A)$ increases, what must be true: $P(A \mid B)$ decreases $P(B \mid A)$ decreases $P(B)$ decreases All of above
$A$ and $B$ are two events. If $P(A, B)$ decreases while $P(A)$ increases, what must be true:$P(A \mid B)$ decreases$P(B \mid A)$ decreases$P(B)$ decreasesAll of above
User Avatar
13 votes
2 answers
106
ISI2015-MMA-7
Suppose $X$ is distributed as Poisson with mean $λ.$ Then $E(1/(X + 1))$ is $\frac{e^{\lambda }-1}{\lambda }$ $\frac{e^{\lambda }-1}{\lambda +1}$ $\frac{1-e^{-\lambda }}{\lambda}$ $\frac{1-e^{-\lambda }}{\lambda + 1}$
Suppose $X$ is distributed as Poisson with mean $λ.$ Then $E(1/(X + 1))$ is$\frac{e^{\lambda }-1}{\lambda }$$\frac{e^{\lambda }-1}{\lambda +1}$$\frac{1-e^{-\lambda }}{\l...
User Avatar
13 votes
2 answers
110
TIFR CSE 2012 | Part A | Question: 9
The probability of throwing six perfect dices and getting six different faces is $1 -\dfrac{ 6!} { 6^{6}}$ $\dfrac{6! }{ 6^{6}}$ $6^{-6}$ $1 - 6^{-6}$ None of the above
The probability of throwing six perfect dices and getting six different faces is$1 -\dfrac{ 6!} { 6^{6}}$$\dfrac{6! }{ 6^{6}}$$6^{-6}$$1 - 6^{-6}$None of the above
User Avatar
Page: