Log In

Recent questions tagged binomial-distribution

0 votes
1 answer
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 146 views
0 votes
0 answers
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 238 views
0 votes
1 answer
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 434 views
0 votes
1 answer
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 167 views
1 vote
1 answer
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 210 views
3 votes
1 answer
11 votes
4 answers
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 1.5k views
21 votes
6 answers
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 2k views
10 votes
2 answers
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
6 answers
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 12.9k views
28 votes
4 answers
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 5k views
11 votes
2 answers
18 votes
3 answers
If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads? $\left(\dfrac{1}{3}\right)$ $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{2}\right)$ $\left(\dfrac{2}{3}\right)$
asked Sep 29, 2014 in Probability jothee 4.5k views
20 votes
9 answers
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 4.2k views
22 votes
7 answers
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 4.5k views
12 votes
5 answers
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is $\frac{1}{16}$ $\frac{1}{8}$ $\frac{7}{8}$ $\frac{15}{16}$
asked Sep 16, 2014 in Probability Kathleen 6.2k views
To see more, click for the full list of questions or popular tags.