A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2 is the variance of X, then 100 σ2 is equal to

Question

A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2 is the variance of X, then 100 σ2 is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/e469eb2caeb03c749cbda45a75a7d919-.png" /> σ2=∑ X2 P(X)-(∑ X P(X))2=(56/100) 100 σ2=56

Q. A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If $σ^{2}$ is the variance of X, then $100 σ^{2}$ is equal to

$σ^{2} = \sum X^{2} P (X) - (\sum XP (X))^{2} = \frac{56}{100}$
$100 σ^{2} = 56$

Q. A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2 is the variance of X, then 100σ2 is equal to

σ2=∑X2P(X)−(∑XP(X))2=10056​ 100σ2=56

Q. A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If $σ^{2}$ is the variance of X, then $100 σ^{2}$ is equal to

$σ^{2} = \sum X^{2} P (X) - (\sum XP (X))^{2} = \frac{56}{100}$
$100 σ^{2} = 56$