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 :

Question

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 : (A) 6 (B) 4 (C) (6/25) (D) (12/5)

Tardigrade Admin · Accepted Answer

n = 10 p(Probability of drawing a green ball) =(15/25) ∴ p=(3/5), q=(2/5) var(X) = n.p.q =10. (6/25)=(12/5)

Q. 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 :

6

4

$\frac{6}{25}$

$\frac{12}{5}$

$n = 10$
p(Probability of drawing a green ball) $= \frac{15}{25}$
$∴ p = \frac{3}{5}, q = \frac{2}{5}$
var(X) = n.p.q
$= 10. \frac{6}{25} = \frac{12}{5}$

Q. 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 :

6

4

256​

512​

n=10 p(Probability of drawing a green ball) =2515​ ∴p=53​,q=52​ var(X) = n.p.q =10.256​=512​

Q. 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 :

$\frac{6}{25}$

$\frac{12}{5}$

$n = 10$
p(Probability of drawing a green ball) $= \frac{15}{25}$
$∴ p = \frac{3}{5}, q = \frac{2}{5}$
var(X) = n.p.q
$= 10. \frac{6}{25} = \frac{12}{5}$