Which one is not a requirement of a binomial distribution? There are 2 outcomes for each trial There is a fixed number of trials The outcomes must be

Monjur Alahi

Which one is not a requirement of a binomial distribution? There are 2 outcomes for each trial There is a fixed number of trials The outcomes must be dependent on each other The probability of successes must be the same for all the trials.
A. There are 2 outcomes for each trial
B. There is a fixed number of trials
C. The outcomes must be dependent on each other
D. The probability of success must be the same for all the trials


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

we know that, in a Binomial distribution,
(i) There are 2 outcomes for each trial.
(ii) There is a fixed number of trials.
(iii) The probability of succes must be the same for all the trials.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
The basic requirement for facilitating the production and distribution of goods and services for economic development.

Social infrastructure

2 Answers 1 views
Collection of outcomes of an experiment is  A) Event  B) Trial  C) Probability  D) Random experiment

Correct option is: A) Event Collection of outcomes of an experiment is event.

2 Answers 1 views
Collection of outcomes of an experiment is  A) Event  B) Trial  C) Probability  D) Random experiment 2 Answers 1 views
State the accounts which are not considered while preparing the Trial Balance by Balance Method Accounts which do not show any balance are not conside

Correct Answer - Balance Method.

2 Answers 1 views
Suppose a random variable X follows the binomial distribution with parameters n and p, where `0<p<1` If `P(X=r)/(P(X=n-r))` is independent of n and r,

`becauseP(X=r)=""^(n)C_(r)(p)^(r )(q)^(n-r)=(n!)/((n-r)!r!)(p)^(r )(1-p)^(n-r)[becauseq=1-p]…(i)` `P(X=0)=(1-p)^(n)` and `P(X=n-r)=""^(n)C_(n-r)(p)^(n-r)(q)^(n-(n-r))` `=(n!)/((n-r)!r!)(p)^(n-r)(1-p^(+r)[becauseq=1-p]because""^(n)C_(r)=""^(n)C_(n-r)]...(ii)` Now, `(P(x=r))/(P(x=n-r))=((n!)/((n-r)!r!)p^(r )(1-p)^(n-r))/((n!)/((n-r)!r!)p^(n-r)(1-p)^(+r))` [using eqs. (i) and (ii)] `=((1-p)/p)^(n-r)xx1/((1-p)/p)^(r)` Above expression is independent of n and r, if `(1-p)/p=1rArr1/p=2rArrp=1/2`

2 Answers 1 views
A sample space consists of 9 elementary outcomes outcomes `E_(1), E_(2)`,…, `E_(9)` whose probabilities are: P(E_(1))=P(E_(2)) = 0.09, P(E_(3))=P(E_(4

(a) `P(A) = P(E_(1)) + P(E_(5)) + P(E_(8))` `=0.09 + 0.1 + 0.06 = 0.25` (b) `P(B) = P(E_(2)) + P(E_(5)) + P(E_(8)) = P(E_(9))` `= 0.09 + 0.1 +...

2 Answers 1 views
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

Correct answer is (b) 5

2 Answers 1 views
In the expansion of `(3^(-x//4)+3^(5x//4))^n` the sum of binomial coefficient is 64 and term with the greatest binomial coefficient exceeds the third

Correct Answer - A To get sum of coefficients put `x = 0` Given that sum of coefficient is `2^(n) = 64` or `n = 6` The greatest binomial coefficeint is...

2 Answers 2 views
An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

Let p be its probability of success and q be the failure. Then p =2q. Also, p+q=1. It gives `p=2//3and q=1//3.` P {4 successes in 6 trials} `=""^(6)C_(4)p^(4)q^(2)` `=""^(6)C_(4)((2)/(3))^(4)((1)/(3))^(2)" "(1)`...

2 Answers 2 views
The probability of happening an event A in one trial is 0.4. Find the probability that the event A happens at least one in three independent trials.

Correct Answer - `4.0.784` Here `P(A)=0.4and P(barA)=0.6` Probability that A does not happen at all is `(0.6)^(3).` Thus, required probability is `1-(0.6)^(3)=0.784.`

2 Answers 1 views
X