Find the binomial probability distribution wh | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a binomial distribution, the mean is given by $E(X) = np = 3$ and the variance is given by $Var(X) = npq = 2$. Dividing the variance by the me

Vedclass · Accepted Answer

$\left(\frac{2}{3}+\frac{1}{3}\right)^9$

Vedclass · Answer

(A) For a binomial distribution, the mean is given by $E(X) = np = 3$ and the variance is given by $Var(X) = npq = 2$. Dividing the variance by the mean, we get $\frac{npq}{np} = \frac{2}{3}$, which implies $q = \frac{2}{3}$. Since $p + q = 1$, we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$. Substituting $p = \frac{1}{3}$ into $np = 3$, we get $n 	imes \frac{1}{3} = 3$, which implies $n = 9$. The binomial distribution is given by $(q + p)^n = \left(\frac{2}{3} + \frac{1}{3}ight)^9$.

Find the binomial probability distribution whose mean is $3$ and variance is $2$.

Similar Questions

$A$ fair die is thrown $4$ times. If getting a prime number on the die is considered as a success, then the probability of getting no success at all is ...

If a fair coin is tossed five times,then the probability of getting exactly three heads is . . . . . . .

If a fair coin is tossed $10$ times,find the probability of getting at least six heads.

For an initial screening of an entrance exam,a candidate is given $50$ problems to solve. If the probability that the candidate can solve any problem is $\frac{4}{5}$,then the probability that he is unable to solve less than $2$ problems is:

Two cards are drawn at random one after the other with replacement from a pack of $52$ playing cards. Then the variance of the random variable of the number of spade cards among the drawn cards is

Vedclass Test Series

Exam Paper Generator

Online Exam Module