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Recent questions tagged binomialdistribution
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1
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Hk Dass
In a binomial distribution the sum and the product of mean and variance are $\Large \frac{25}{3}$ and $\Large \frac{50}{3}$ respectively. The distribution is _______. Note : I've not included the options to avoid KBC in comments
asked
Aug 31
in
Probability
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Mk Utkarsh
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binomialdistribution
probability
0
votes
1
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2
Probability
We have applied Bernoulli equation to solve the answer. But, why the answer isn't C(90,5)÷C(100,5)?
asked
Aug 9
in
Probability
by
Arjun045
(
9
points)

31
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binomialdistribution
probability
+10
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3
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3
GATE2018GA10
A six sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the most likely outcome of the experiment? Three green faces and four red faces. Four green faces and three red faces. Five green faces and two red faces. Six green faces and one red face
asked
Feb 14
in
Numerical Ability
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gatecse
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gate2018
numericalability
probability
binomialdistribution
normal
+1
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1
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4
$a_n = 4^n + 6^n$
If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$
asked
May 19, 2017
in
Set Theory & Algebra
by
Debashish Deka
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57.6k
points)

120
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binomialdistribution
+2
votes
1
answer
5
Hashing+Probaility
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Oct 8, 2016
in
DS
by
Rahul Jain25
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11.6k
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237
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hashing
probability
uniformhashing
binomialdistribution
+6
votes
2
answers
6
TIFR2011A3
The probability of three consecutive heads in four tosses of a fair coin is. $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{8}\right)$ $\left(\dfrac{1}{16}\right)$ $\left(\dfrac{3}{16}\right)$ $\text{None of the above.}$
asked
Oct 17, 2015
in
Probability
by
makhdoom ghaya
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40.5k
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341
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tifr2011
probability
binomialdistribution
+8
votes
4
answers
7
TIFR2010B38
Suppose three coins are lying on a table, two of them with heads facing up and one with tails facing up. One coin is chosen at random and flipped. What is the probability that after the flip the majority of the coins(i.e., at least two of them) will have heads facing up? ... $\left(\frac{1}{4}\right)$ $\left(\frac{1}{4}+\frac{1}{8}\right)$ $\left(\frac{2}{3}\right)$
asked
Oct 11, 2015
in
Probability
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makhdoom ghaya
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40.5k
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703
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tifr2010
probability
binomialdistribution
+7
votes
1
answer
8
TIFR2010A6
Given 10 tosses of a coin with probability of head = .$4$ = ($1$  the probability of tail), the probability of at least one head is? $(.4)^{10}$ $1  (.4)^{10}$ $1  (.6)^{10}$ $(.6)^{10}$ $10(.4) (.6)^{9}$
asked
Oct 2, 2015
in
Probability
by
makhdoom ghaya
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40.5k
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327
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tifr2010
probability
binomialdistribution
+28
votes
4
answers
9
GATE2005IT32
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is $3$ $4$ $5$ $6$
asked
Nov 3, 2014
in
Probability
by
Ishrat Jahan
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19.1k
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3.9k
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gate2005it
probability
binomialdistribution
expectation
normal
+16
votes
3
answers
10
GATE2006IT22
When a coin is tossed, the probability of getting a Head is $p, 0 < p < 1$. Let $N$ be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of $N$ is $\dfrac{1}{p}$ $\dfrac{1}{(1  p)}$ $\dfrac{1}{p^{2}}$ $\dfrac{1}{(1  p^{2})}$
asked
Oct 31, 2014
in
Probability
by
Ishrat Jahan
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19.1k
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1.6k
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gate2006it
probability
binomialdistribution
expectation
normal
+10
votes
5
answers
11
GATE200552
A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. The probability that two such randomly generated strings are not identical is: $\frac{1}{2^n}$ $1  \frac{1}{n}$ $\frac{1}{n!}$ $1  \frac{1}{2^n}$
asked
Sep 21, 2014
in
Probability
by
gatecse
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18.3k
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1.3k
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gate2005
probability
binomialdistribution
easy
+11
votes
4
answers
12
GATE200621
For each element in a set of size $2n$, an unbiased coin is tossed. The $2n$ coin tosses are independent. An element is chosen if the corresponding coin toss was a head. The probability that exactly $n$ elements are chosen is $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$
asked
Sep 17, 2014
in
Probability
by
Rucha Shelke
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1.7k
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gate2006
probability
binomialdistribution
normal
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