# Recent questions tagged binomial-distribution 0 votes
1 answer
1
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.
0 votes
0 answers
2
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 ?
0 votes
1 answer
3
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
0 votes
1 answer
4
We have applied Bernoulli equation to solve the answer. But, why the answer isn't C(90,5)÷C(100,5)?
1 vote
1 answer
5
If $a_n = 4^n + 6^n$ Find the value of $a_{40} \text { mod } 25$
3 votes
1 answer
6
11 votes
4 answers
7
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
21 votes
6 answers
8
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? ... $\left(\frac{1}{4}\right)$ $\left(\frac{1}{4}+\frac{1}{8}\right)$ $\left(\frac{2}{3}\right)$
10 votes
2 answers
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? $(.4)^{10}$ $1 - (.4)^{10}$ $1 - (.6)^{10}$ $(.6)^{10}$ $10(.4) (.6)^{9}$
48 votes
6 answers
10
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 $3$ $4$ $5$ $6$
28 votes
4 answers
11
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 $\dfrac{1}{p}$ $\dfrac{1}{(1 - p)}$ $\dfrac{1}{p^{2}}$ $\dfrac{1}{(1 - p^{2})}$
11 votes
2 answers
12
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 $\dfrac{1}{4}.$ ... $\left(\dfrac{1}{2}\right)$ $\left(\dfrac{7}{16}\right)$ $\left(\dfrac{5}{32}\right)$
18 votes
3 answers
13
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? $\left(\dfrac{1}{3}\right)$ $\left(\dfrac{1}{4}\right)$ $\left(\dfrac{1}{2}\right)$ $\left(\dfrac{2}{3}\right)$
20 votes
9 answers
14
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: $\frac{1}{2^n}$ $1 - \frac{1}{n}$ $\frac{1}{n!}$ $1 - \frac{1}{2^n}$
22 votes
7 answers
15
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 $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$
12 votes
5 answers
16
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is $\frac{1}{16}$ $\frac{1}{8}$ $\frac{7}{8}$ $\frac{15}{16}$
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