Recent questions tagged binomial-distribution

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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.

asked Jan 9 in Mathematical Logic by junaid ahmad Loyal (9.4k points) | 20 views

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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 ?

asked Dec 18, 2018 in Mathematical Logic by shreyansh jain Active (2.2k points) | 46 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

asked Aug 31, 2018 in Probability by Mk Utkarsh Boss (34.8k points) | 83 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)?

asked Aug 9, 2018 in Probability by Arjun045 (23 points) | 34 views

+10 votes

3 answers

5

GATE2018-GA-10
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, 2018 in Numerical Ability by gatecse Boss (18.2k points) | 4.1k views

+1 vote

1 answer

6

$a_n = 4^n + 6^n$
If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$

asked May 19, 2017 in Set Theory & Algebra by Debashish Deka Veteran (58.2k points) | 126 views

+2 votes

1 answer

7

Hashing+Probaility

asked Oct 8, 2016 in DS by Rahul Jain25 Boss (11.7k points) | 249 views

+6 votes

2 answers

8

TIFR2011-A-3
The probability of three consecutive heads in four tosses of a fair coin is. $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{8}\right)$ $\left(\dfrac{1}{16}\right)$ $\left(\dfrac{3}{16}\right)$ $\text{None of the above.}$

asked Oct 17, 2015 in Probability by makhdoom ghaya Boss (41.1k points) | 394 views

+9 votes

5 answers

9

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? ... $\left(\frac{1}{4}\right)$ $\left(\frac{1}{4}+\frac{1}{8}\right)$ $\left(\frac{2}{3}\right)$

asked Oct 11, 2015 in Probability by makhdoom ghaya Boss (41.1k points) | 799 views

+7 votes

1 answer

10

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? $(.4)^{10}$ $1 - (.4)^{10}$ $1 - (.6)^{10}$ $(.6)^{10}$ $10(.4) (.6)^{9}$

asked Oct 2, 2015 in Probability by makhdoom ghaya Boss (41.1k points) | 372 views

+32 votes

4 answers

11

GATE2005-IT-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 $3$ $4$ $5$ $6$

asked Nov 3, 2014 in Probability by Ishrat Jahan Boss (19.1k points) | 4.6k views

+19 votes

3 answers

12

GATE2006-IT-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 $\dfrac{1}{p}$ $\dfrac{1}{(1 - p)}$ $\dfrac{1}{p^{2}}$ $\dfrac{1}{(1 - p^{2})}$

asked Oct 31, 2014 in Probability by Ishrat Jahan Boss (19.1k points) | 1.9k views

+15 votes

5 answers

13

GATE2005-52
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: $\frac{1}{2^n}$ $1 - \frac{1}{n}$ $\frac{1}{n!}$ $1 - \frac{1}{2^n}$

asked Sep 21, 2014 in Probability by gatecse Boss (18.2k points) | 1.6k views

+14 votes

4 answers

14

GATE2006-21
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 $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$

asked Sep 17, 2014 in Probability by Rucha Shelke Active (3.7k points) | 1.9k views

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