Two balls are drawn from an urn containing 7 | vedclass

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(D) Total number of balls = $7 + 6 + 8 = 21$. Number of white balls = $7$. Number of non-white balls = $6 + 8 = 14$. We need to find the probability t

Vedclass · Accepted Answer

$\frac{17}{30}$

Vedclass · Answer

(D) Total number of balls = $7 + 6 + 8 = 21$. Number of white balls = $7$. Number of non-white balls = $6 + 8 = 14$. We need to find the probability that at least one ball is white. It is easier to calculate the probability of the complement event: $P(	ext{at least one white}) = 1 - P(	ext{no white balls})$. If no white balls are drawn,both balls must be non-white. Probability of drawing the first non-white ball = $\frac{14}{21}$. After drawing one non-white ball,$13$ non-white balls remain out of $20$ total balls. Probability of drawing the second non-white ball = $\frac{13}{20}$. $P(	ext{no white}) = \frac{14}{21} 	imes \frac{13}{20} = \frac{2}{3} 	imes \frac{13}{20} = \frac{26}{60} = \frac{13}{30}$. Therefore,$P(	ext{at least one white}) = 1 - \frac{13}{30} = \frac{17}{30}$.

Two balls are drawn from an urn containing $7$ white,$6$ red,and $8$ black balls one after the other without replacement. The probability that at least one of them is white is:

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