Question

If a bag contains 6 red balls, 4 green balls, and 3 blue balls. If we choose a ball, then another ball without putting the first one in the bag, what is the probability that the first ball is green anda the second one red

Answer 1

hello,

red 6

green 4

blue 3

• first ball is green:

$\frac{4}{13}$

• second ball is red:

$\frac{6}{12} = \frac{1}{2}$

The probability that these two events happen at the same time is:

$\frac{4}{13} \times \frac{1}{2} = \frac{4}{26} = \frac{2}{13}$

$\boxed{crsjr}$