Marbles - Probability?

Question

A box contains 2 green marbles and 4 white marbles.

A. Two marbles are drawn , but the first is not replaced before drawing the second. What is the probability that both are green?

B. Two marbles are drawn without replacement. What is the probability that they are the same color?

C. Two marbles are drawn without replacement. What is the probability that they are not the same color?

Answer 1

Ok, let me see if I can help.

We have a total of 6 marbles, 2 green and 4 white.

a) The first marble picked has a 2/6 prob. of being green. Since it is not replaced, or put back into the bag, the number of marbles is now 5. Since we picked the first one to be green, which gives the 2/6 prob. the prob. of getting the second one green is 1/5. If we multiply these to probs. together we get (2/6)(1/5)=(1/15)

b) Either we have 2 green or 2 white. For green we have the same answer as above which is (1/15). For white we have a 4/6 prob. on the first pick and (3/5) on the second. This give the prob. of picking 2 white marbles as (4/6)(3/5)=(2/5). To find the prob. of picking both marbles of the same color, we ADD (1/15) and (2/5), giving (7/15).

c) Notice in b)that we found the prob. of picking 2 matching marbles as (7/15) so if we treat the prob. of picking ANY 2 marbles is 1, we can find the prob. of NOT picking 2 matching marbles. Since the prob. of picking ANY 2 marbles is the prob. of picking 2 matching (7/15) PLUS the prob. of picking 2 NON-matching marbles is equal to 1 we get:

(15/15)-(7/15)=8/15, this is the answer.

Answer 2

There are 6 marbles total

A. P(green) = 2/6, now there are 5 marbles remaining and one is green
P(green again) = 1/5
P(both green) = 2/6 * 1/5 = 1/15

B. You know the probability they are both green from A, so you need prob both are blue
P(blue) = 4/6 and P(blue again) = 3/5
P(both blue) = 4/6*3/5 = 2/5

P(both same color) = P(both green) or P(both blue)
=P(both green) + P(both blue) = 1/15 + 2/5 = 7/15

C. P(one each color) = P(green)P(blue) = (2/6)*(4/5) = 4/15
Also it could be P(blue)P(green) = (4/6)*(2/5) = 4/15

So in total the probability is the sum of those = 8/15

Answer 3

a) The probability of drawing the first green marble is 2 out of 6. The probability of drawing the second green one is 1 out of 5. So the probablility of drawing them both is (2/6)(1/5) = 1/15.

b) The probablility of drawing a white marble is 4 out of 6. The probablility of drawing a second white marble is 3 out of 5. So the probability of drawing them both is (4/6)(3/5) = 2/5. The probability of drawing 2 greens or whites is (1/15)+(2/5) = 7/15.

c) The probability of them being different colors is just 1 - the probability of them being the same color. 1 - (7/15) = 8/15.

Answer 4

A. probability that first one is green is 2/6. probability that second one is green is 1/5 total prob = 2/30

B. Probability that both are green = 2/30. Probability that both are white = 4/6*3/5 = 12/30. So probability that both are of the same color is 14/30

C 1-14/30 = 16/30 = 8/15

Answer 5

There are 6 marbles entire A. P(eco-friendly) = 2/6, now there are 5 marbles terrific and one is eco-friendly P(eco-friendly decrease back) = a million/5 P(both eco-friendly) = 2/6 * a million/5 = a million/15 B. you comprehend the threat they're both eco-friendly from A, so that you opt for prob both are blue P(blue) = 4/6 and P(blue decrease back) = 3/5 P(both blue) = 4/6*3/5 = 2/5 P(both similar colour) = P(both eco-friendly) or P(both blue) =P(both eco-friendly) + P(both blue) = a million/15 + 2/5 = 7/15 C. P(one each colour) = P(eco-friendly)P(blue) = (2/6)*(4/5) = 4/15 also it would want to opt for to be P(blue)P(eco-friendly) = (4/6)*(2/5) = 4/15 So in entire the threat is the sum of those = 8/15

Answer 6

a is 1/15..there are 15 combinations and only one is green green

b is 7/15...out of the 15 combinations, 8 are white/green combos

c is 8/15....out of the 15 combinations, 8 are white/green combos