A box contains 15 green and 10 yellow balls. | vedclass

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

Question

(D) The number of green balls drawn follows a binomial distribution $B(n, p)$.Here,the total number of balls is $15 + 10 = 25$.The probability of draw

Vedclass · Accepted Answer

$\frac{12}{5}$

Vedclass · Answer

(D) The number of green balls drawn follows a binomial distribution $B(n, p)$.Here,the total number of balls is $15 + 10 = 25$.The probability of drawing a green ball in a single trial is $p = \frac{15}{25} = \frac{3}{5}$.The probability of drawing a yellow ball is $q = 1 - p = \frac{10}{25} = \frac{2}{5}$.The number of trials is $n = 10$.The variance of a binomial distribution is given by $\sigma^2 = npq$.Substituting the values,we get $\sigma^2 = 10 	imes \frac{3}{5} 	imes \frac{2}{5} = 10 	imes \frac{6}{25} = \frac{60}{25} = \frac{12}{5}$.

$A$ box contains $15$ green and $10$ yellow balls. If $10$ balls are randomly drawn,one by one,with replacement,then the variance of the number of green balls drawn is

Similar Questions

The probability distribution of the number of doublets in four throws of a pair of dice is given by

If the sum of mean and variance of a binomial distribution for $5$ trials is $1.8$,then the probability of success is:

$A$ person can kill a bird with a probability of $3/4$. He tries $5$ times. What is the probability that he may not kill the bird at all?

The sum and product of the mean and variance of a binomial distribution are $82.5$ and $1350$ respectively. The number of trials in the binomial distribution is:

In a binomial distribution,the probability of success is $\frac{1}{4}$ and the standard deviation is $3$. Find its mean.

Vedclass Test Series

Exam Paper Generator

Online Exam Module