# QOTD-A bag contains 5 red, 8 blue, and 3 green balls. If two balls are drawn in such a way that the second ball is drawn without replacing the first ball in the bag, then what is the probability that the first ball is blue ball and second ball is a green

2019-08-03 | Team PendulumEdu

Options:

A. $$\frac{7}{10}$$

B. $$\frac{11}{24}$$

C. $$\frac{1}{10}$$

D. $$\frac{1}{5}$$

Solution:

According to the question, a bag contains 5 red, 8 blue, and 3 green balls

So,

Total number of balls = 5 + 8 + 3 = 16 balls

Two balls are drawn in such a way that the second ball is drawn without replacing the first ball in the bag

Thus, the total number of ways for taking out the first ball as a blue ball (out of 8 blue balls) = $$^{8}C_{1}$$ = 8

Total number of ways for taking out the first ball out of 16 balls = $$^{16}C_{1}$$ = 16

And, the total number of ways for taking out the second ball as Green ball (out of 3 green balls) = $$^{3}C_{1}$$ = 3

Total number of ways for taking out the second ball out of 15 balls = $$^{15}C_{1}$$ = 15

As we know,

probability=(favorable outcomes)/(Total outcomes)

So, the probability of the first ball being a blue ball and the second ball being green ball is

$$=\frac{8}{16}*\frac{3}{15}=\frac{1}{2}*\frac{1}{5}=\frac{1}{10}$$

Hence, C is the correct answer.

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