A bag contains 7 green and 5 black balls. 3 b | vedclass

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(D) Total number of balls $= 7 + 5 = 12$. Probability of drawing the first green ball $= \frac{7}{12}$. Since the balls are not replaced,the number of

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$\frac{7}{44}$

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(D) Total number of balls $= 7 + 5 = 12$. Probability of drawing the first green ball $= \frac{7}{12}$. Since the balls are not replaced,the number of green balls remaining is $6$ and the total number of balls remaining is $11$. Probability of drawing the second green ball $= \frac{6}{11}$. After drawing two green balls,the number of green balls remaining is $5$ and the total number of balls remaining is $10$. Probability of drawing the third green ball $= \frac{5}{10} = \frac{1}{2}$. The probability that all three balls are green is the product of these individual probabilities: $P = \frac{7}{12} 	imes \frac{6}{11} 	imes \frac{5}{10} = \frac{7 	imes 6 	imes 5}{12 	imes 11 	imes 10} = \frac{210}{1320} = \frac{7}{44}$.

$A$ bag contains $7$ green and $5$ black balls. $3$ balls are drawn at random one after the other. If the balls are not replaced,then the probability of all three balls being green is

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