Answer:
There are 18 balls in the bag:
6 are red
6 are yellow
6 are green.
A ball is drawn at random and then returned.
Then another ball is drawn.
Now, let's analyze each option and let's see if it is true or not.
A) The probability of drawing at random a green ball is equal to the quotient between the number of green balls and the total number of balls.
Then for the first draw, the probability is 6/18
Because the drawn ball is returned, for the second draw we still have 6 green balls and 18 balls in total, then the probability still is 6/18
Statement A is true
B) if we define the notation:
YR
as:
Yellow in the first draw, red in the second draw.
The possible outcomes are:
YR
YG
YY
RR
RG
RY
GR
GG
GY
So we have a total of 9 outcomes, we also can calculate this in the following way:
For the first event (first draw) we have 3 possible outcomes (Y, G, and R)
For the second event (second draw) we have again 3 possible outcomes.
The total number of outcomes is the product between the possible outcomes for each event, this is:
3*3 = 9
Then statement B is false.
C) The probability of drawing a red ball first is:
p = 6/18
The probability of NOT drawing a red ball second, is equal to the probability of drawing a yellow or green ball second (we have 12 balls that are green or yellow, and 18 total balls) then this probability is:
q = 12/18
The probabilities are different, then we can conclude that statement C is false.
fD) The expected probabilty of drawing a red ball and then a yellow ball is the product of the individual probabilities, this is:
prob of drawing a red ball first = 6/18
prob of drawing a yellow ball second = 6/18
Joint probability:
P = (6/18)*(6/18) = (1/3)*(1/3) = 1/9
So statement D is true. (it says 19, i suppose that there is a mistake in this sentence and it actually should say 1/9)