8. a jar contains four blue marbles and two red marbles. suppose you choose a marble at random, and do not replace it. then you choose a second marble. find the probability of each event to the nearest tenth of a percent. a) the first marble is blue and the second is red. b) both marbles are blue. ( in advance and show your work so i know how you got your answer! : d )

(a) To find the probability that the first marble is red and the second in blue we need to consider the probability of each. First as there are 4 Blue marbles and 2 Red ones, there are 6 marbles all together. The probability that the first marble is Blue is . That is, there are 4 blue marbles out of 6 total.

Since the blue marble you selected is not replaced. There are now 3 blue marbles and 2 red ones. That is, 5 total. The probability that the second marble is red is .

We multiply these to obtain the probability of Blue followed by Red. The probability is

Typically you leave the probability as a fraction but since you were asked for the answer to the nearest tenth of a percent we divide 8 by 30 to obtain .266666 which we multiply by 100 to get 26.66666% and to one decimal place that gives the answer 26.7%

(b) To find the probability that both marbles are blue we need to find probability of each. First as there are 4 Blue marbles and 2 Red ones, there are 6 marbles all together. The probability that the first marble is Blue is . That is, there are 4 blue marbles out of 6 total.

Since the blue marble you selected is not replaced. There are now 3 blue marbles and 2 red ones. That is, 5 total. The probability that the second marble is also blue is is .

We multiply these to obtain the probability of Blue followed by Blue. The probability is

Typically you leave the probability as a fraction but since you were asked for the answer to the nearest tenth of a percent we divide 12 by 30 to obtain .4 which we multiply by 100 to get 40%