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

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(B) The problem follows a binomial distribution because the balls are drawn with replacement,making each trial independent.Here,the total number of tr

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$\frac{12}{5}$

Vedclass · Answer

(B) The problem follows a binomial distribution because the balls are drawn with replacement,making each trial independent.Here,the total number of trials is $n = 10$.The probability of drawing a green ball in a single trial is $p = \frac{15}{15 + 10} = \frac{15}{25} = \frac{3}{5}$.The probability of not drawing a green ball (drawing a yellow ball) is $q = 1 - p = 1 - \frac{3}{5} = \frac{2}{5}$.The variance of a binomial distribution is given by the formula $	ext{Variance} = npq$.Substituting the values: $	ext{Variance} = 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:

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