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

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

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

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

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

3 votes

1 answer

6

Hashing+Probaility

asked Oct 8, 2016 in DS Rahul Jain25 358 views

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

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

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

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

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

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

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

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