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Recent questions tagged binomial-distribution
0
votes
1
answer
1
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.
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 anything about a replacement. So is there any by default approach that we have to use Binomial if nothing is a mention about a replacement.
asked
Jan 9, 2019
in
Mathematical Logic
junaid ahmad
134
views
made-easy-test-series
probability
binomial-distribution
0
votes
0
answers
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 ?
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 ?
asked
Dec 18, 2018
in
Mathematical Logic
shreyansh jain
226
views
binomial-distribution
0
votes
1
answer
3
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
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, 2018
in
Probability
Mk Utkarsh
403
views
binomial-distribution
probability
0
votes
1
answer
4
Probability
We have applied Bernoulli equation to solve the answer. But, why the answer isn't C(90,5)÷C(100,5)?
We have applied Bernoulli equation to solve the answer. But, why the answer isn't C(90,5)÷C(100,5)?
asked
Aug 9, 2018
in
Probability
Arjun045
149
views
binomial-distribution
probability
1
vote
1
answer
5
$a_n = 4^n + 6^n$
If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$
If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$
asked
May 19, 2017
in
Set Theory & Algebra
dd
202
views
binomial-distribution
3
votes
1
answer
6
Hashing+Probaility
asked
Oct 8, 2016
in
DS
Rahul Jain25
358
views
hashing
probability
uniform-hashing
binomial-distribution
11
votes
4
answers
7
TIFR2011-A-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 $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{8}\right)$ $\left(\dfrac{1}{16}\right)$ $\left(\dfrac{3}{16}\right)$ None of the above
asked
Oct 17, 2015
in
Probability
makhdoom ghaya
1.4k
views
tifr2011
probability
binomial-distribution
20
votes
6
answers
8
TIFR2010-B-38
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? ... $\left(\frac{1}{4}\right)$ $\left(\frac{1}{4}+\frac{1}{8}\right)$ $\left(\frac{2}{3}\right)$
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
makhdoom ghaya
1.9k
views
tifr2010
probability
binomial-distribution
10
votes
2
answers
9
TIFR2010-A-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? $(.4)^{10}$ $1 - (.4)^{10}$ $1 - (.6)^{10}$ $(.6)^{10}$ $10(.4) (.6)^{9}$
asked
Oct 2, 2015
in
Probability
makhdoom ghaya
1.2k
views
tifr2010
probability
binomial-distribution
48
votes
5
answers
10
GATE IT 2005 | Question: 32
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 $3$ $4$ $5$ $6$
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
Ishrat Jahan
11.6k
views
gate2005-it
probability
binomial-distribution
expectation
normal
27
votes
4
answers
11
GATE IT 2006 | Question: 22
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 $\dfrac{1}{p}$ $\dfrac{1}{(1 - p)}$ $\dfrac{1}{p^{2}}$ $\dfrac{1}{(1 - p^{2})}$
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
Ishrat Jahan
4.8k
views
gate2006-it
probability
binomial-distribution
expectation
normal
20
votes
9
answers
12
GATE CSE 2005 | Question: 52
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: $\frac{1}{2^n}$ $1 - \frac{1}{n}$ $\frac{1}{n!}$ $1 - \frac{1}{2^n}$
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
gatecse
4k
views
gate2005-cse
probability
binomial-distribution
easy
20
votes
6
answers
13
GATE CSE 2006 | Question: 21
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 $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$
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
Rucha Shelke
4.3k
views
gate2006-cse
probability
binomial-distribution
normal
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