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Recent questions tagged binomial-distribution
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Best Open Video Playlist for Binomial Distributions Topic | Probability
Please list out the best free available video playlist for Binomial Distributions Topic from Probability as an answer here (only one playlist per answer). We'll then select the best playlist and add to GO classroom video lists ... ones are more likely to be selected as best. For the full list of selected videos please see here
makhdoom ghaya
asked
in
Study Resources
Aug 15
by
makhdoom ghaya
12
views
missing-videos
go-classroom
free-videos
video-links
binomial-distribution
0
votes
1
answer
2
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.
junaid ahmad
asked
in
Mathematical Logic
Jan 9, 2019
by
junaid ahmad
277
views
made-easy-test-series
probability
binomial-distribution
0
votes
0
answers
3
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 ?
shreyansh jain
asked
in
Mathematical Logic
Dec 18, 2018
by
shreyansh jain
401
views
binomial-distribution
1
vote
1
answer
4
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
Mk Utkarsh
asked
in
Probability
Aug 31, 2018
by
Mk Utkarsh
676
views
binomial-distribution
probability
1
vote
1
answer
5
Probability
We have applied Bernoulli equation to solve the answer. But, why the answer isn't C(90,5)÷C(100,5)?
Arjun045
asked
in
Probability
Aug 9, 2018
by
Arjun045
358
views
binomial-distribution
probability
1
vote
1
answer
6
$a_n = 4^n + 6^n$
If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$
dd
asked
in
Set Theory & Algebra
May 19, 2017
by
dd
289
views
binomial-distribution
3
votes
1
answer
7
Hashing+Probaility
Rahul Jain25
asked
in
DS
Oct 8, 2016
by
Rahul Jain25
490
views
hashing
probability
uniform-hashing
binomial-distribution
13
votes
4
answers
8
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
makhdoom ghaya
asked
in
Probability
Oct 17, 2015
by
makhdoom ghaya
2.1k
views
tifr2011
probability
binomial-distribution
23
votes
6
answers
9
TIFR CSE 2010 | Part B | Question: 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 ... $\left(\frac{1}{4}+\frac{1}{8}\right)$ $\left(\frac{2}{3}\right)$
makhdoom ghaya
asked
in
Probability
Oct 11, 2015
by
makhdoom ghaya
2.7k
views
tifr2010
probability
binomial-distribution
12
votes
2
answers
10
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? $(.4)^{10}$ $1 - (.4)^{10}$ $1 - (.6)^{10}$ $(.6)^{10}$ $10(.4) (.6)^{9}$
makhdoom ghaya
asked
in
Probability
Oct 2, 2015
by
makhdoom ghaya
1.6k
views
tifr2010
probability
binomial-distribution
60
votes
7
answers
11
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 $3$ $4$ $5$ $6$
Ishrat Jahan
asked
in
Probability
Nov 3, 2014
by
Ishrat Jahan
17.0k
views
gateit-2005
probability
binomial-distribution
expectation
normal
34
votes
4
answers
12
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 $\dfrac{1}{p}$ $\dfrac{1}{(1 - p)}$ $\dfrac{1}{p^{2}}$ $\dfrac{1}{(1 - p^{2})}$
Ishrat Jahan
asked
in
Probability
Oct 31, 2014
by
Ishrat Jahan
6.6k
views
gateit-2006
probability
binomial-distribution
expectation
normal
16
votes
3
answers
13
GATE IT 2007 | Question: 1
Suppose there are two coins. The first coin gives heads with probability $\dfrac{5}{8}$ when tossed, while the second coin gives heads with probability $\dfrac{1}{4}.$ One of the two coins is picked up at random with equal probability and tossed. What is the probability of ... $\left(\dfrac{1}{2}\right)$ $\left(\dfrac{7}{16}\right)$ $\left(\dfrac{5}{32}\right)$
Ishrat Jahan
asked
in
Probability
Oct 30, 2014
by
Ishrat Jahan
2.5k
views
gateit-2007
probability
normal
binomial-distribution
25
votes
9
answers
14
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: $\frac{1}{2^n}$ $1 - \frac{1}{n}$ $\frac{1}{n!}$ $1 - \frac{1}{2^n}$
gatecse
asked
in
Probability
Sep 21, 2014
by
gatecse
5.5k
views
gatecse-2005
probability
binomial-distribution
easy
31
votes
7
answers
15
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 $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$
Rucha Shelke
asked
in
Probability
Sep 17, 2014
by
Rucha Shelke
6.0k
views
gatecse-2006
probability
binomial-distribution
normal
17
votes
5
answers
16
GATE CSE 2002 | Question: 2.16
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is $\frac{1}{16}$ $\frac{1}{8}$ $\frac{7}{8}$ $\frac{15}{16}$
Kathleen
asked
in
Probability
Sep 16, 2014
by
Kathleen
8.6k
views
gatecse-2002
probability
easy
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