A bag contain 8 blue and 12 green balls. Two balls are drawn in succession without replacement. The probability that first is blue and second is…

The bag contains 8 blue balls and 12 green balls, for a total of 20 balls. Two balls are drawn in succession without replacement.

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For the first draw, there are 20 balls total and 8 are blue.

$$P(\text{1st Blue}) = \frac{8}{20}$$P(1st Blue)=208​

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After drawing one blue ball, 19 balls remain in the bag. The 12 green balls are still in the bag.

$$P(\text{2nd Green | 1st Blue}) = \frac{12}{19}$$P(2nd Green | 1st Blue)=1912​

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For both events to occur, multiply the probabilities:

$$P(\text{Blue then Green}) = \frac{8}{20} \times \frac{12}{19}$$P(Blue then Green)=208​×1912​

$$= \frac{96}{380}$$=38096​

$$= \frac{24}{95}$$=9524​

Therefore, the probability is $\frac{24}{95}$9524​.