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Recent questions tagged binomialdistribution
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MadeEasy Test Series: Probability
How to get the idea that we have to use Binomial distribution or Hypergeometric Distribution. I know that if the probability is not changing(i.e with replacement) then we go Binomial otherwise Hypergeometric. But in question, it is not indicating ... So is there any by default approach that we have to use Binomial if nothing is a mention about a replacement.
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Jan 9, 2019
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Mathematical Logic
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junaid ahmad
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madeeasytestseries
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binomialdistribution
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2
Binomial distribution
Is there any relation between MEAN, VARIANCE and MODE for binomial distribution? Let, Mean = 8, variance = 6 for any binomial distribution. np = 8 and npq = 6 => q=$3/4$, p=$1/4$ Now is there any relation to find value of MODE ?
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Dec 18, 2018
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Mathematical Logic
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shreyansh jain
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binomialdistribution
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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
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Aug 31, 2018
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Probability
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Mk Utkarsh
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binomialdistribution
probability
0
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1
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4
Probability
We have applied Bernoulli equation to solve the answer. But, why the answer isn't C(90,5)÷C(100,5)?
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Aug 9, 2018
in
Probability
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Arjun045
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19
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48
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binomialdistribution
probability
+1
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1
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5
$a_n = 4^n + 6^n$
If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$
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May 19, 2017
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Set Theory & Algebra
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dd
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binomialdistribution
+2
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1
answer
6
Hashing+Probaility
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Oct 8, 2016
in
DS
by
Rahul Jain25
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hashing
probability
uniformhashing
binomialdistribution
+6
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2
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7
TIFR2011A3
The probability of three consecutive heads in four tosses of a fair coin is. $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{8}\right)$ $\left(\dfrac{1}{16}\right)$ $\left(\dfrac{3}{16}\right)$ $\text{None of the above.}$
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Oct 17, 2015
in
Probability
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makhdoom ghaya
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tifr2011
probability
binomialdistribution
+13
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6
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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? ... $\left(\frac{1}{4}\right)$ $\left(\frac{1}{4}+\frac{1}{8}\right)$ $\left(\frac{2}{3}\right)$
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Oct 11, 2015
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Probability
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makhdoom ghaya
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tifr2010
probability
binomialdistribution
+7
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1
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9
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? $(.4)^{10}$ $1  (.4)^{10}$ $1  (.6)^{10}$ $(.6)^{10}$ $10(.4) (.6)^{9}$
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Oct 2, 2015
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Probability
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makhdoom ghaya
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tifr2010
probability
binomialdistribution
+37
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4
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10
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 $3$ $4$ $5$ $6$
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Nov 3, 2014
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Probability
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Ishrat Jahan
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gate2005it
probability
binomialdistribution
expectation
normal
+21
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4
answers
11
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 $\dfrac{1}{p}$ $\dfrac{1}{(1  p)}$ $\dfrac{1}{p^{2}}$ $\dfrac{1}{(1  p^{2})}$
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Oct 31, 2014
in
Probability
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Ishrat Jahan
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gate2006it
probability
binomialdistribution
expectation
normal
+16
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7
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12
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: $\frac{1}{2^n}$ $1  \frac{1}{n}$ $\frac{1}{n!}$ $1  \frac{1}{2^n}$
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Sep 21, 2014
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Probability
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gatecse
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gate2005
probability
binomialdistribution
easy
+15
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4
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13
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 $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$
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Sep 17, 2014
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Probability
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Rucha Shelke
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gate2006
probability
binomialdistribution
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