A bag contains 2 white,3 green,and 5 red ball | vedclass

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(D) Total number of balls $= 2 + 3 + 5 = 10$. Let $R_3$ be the event that the third ball drawn is red. By the symmetry of drawing balls without replac

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$\frac{1}{2}$

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(D) Total number of balls $= 2 + 3 + 5 = 10$. Let $R_3$ be the event that the third ball drawn is red. By the symmetry of drawing balls without replacement,the probability that the $k$-th ball drawn is of a certain color is equal to the initial proportion of that color in the bag. Specifically,for any position $k$ (where $1 \le k \le 10$),the probability that the $k$-th ball is red is given by $P(R_k) = \frac{	ext{Number of red balls}}{	ext{Total number of balls}}$. Thus,$P(R_3) = \frac{5}{10} = \frac{1}{2}$. Alternatively,using the law of total probability: $P(R_3) = P(R_3|R_1 R_2)P(R_1 R_2) + P(R_3|R_1 R_2^c)P(R_1 R_2^c) + P(R_3|R_1^c R_2)P(R_1^c R_2) + P(R_3|R_1^c R_2^c)P(R_1^c R_2^c) = \frac{1}{2}$.

$A$ bag contains $2$ white,$3$ green,and $5$ red balls. If three balls are drawn one after the other without replacement,then the probability that the last ball drawn was red is

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