An urn contains 7 red marbles labeled {1,2,3,4,5,6,7} and 5 green marbles labeled {1,2,3,4,5}. Four marbles are pulled out at once. What is the probability that
A. All four marbles are red?
B. More of the marbles are green than red?
C. Both red and green marbles are present?
D. Two of the marbles chosen are both labeled "5"?
I've tried but i need help =/
You are randomly drawing marbles from a bag containing 4 blue and 2 green marbles, replacing the marbles between draws. Which of the following will give the probability of drawing a blue marble on both your first and second draws?
To find the probabilities, we need to calculate the number of favorable outcomes and the total number of possible outcomes.
A. All four marbles are red:
There are 7 red marbles and 12 marbles in total. To calculate the probability of selecting four red marbles, we use the combination formula:
C(7,4) / C(12,4)
where C(n, r) denotes the combination of choosing r objects from a set of n objects.
C(7,4) = 7! / (4!(7-4)!) = 7! / (4!*3!) = (7*6*5) / (3*2*1) = 35
C(12,4) = 12! / (4!(12-4)!) = 12! / (4!*8!) = (12*11*10*9) / (4*3*2*1) = 495
So, the probability of selecting all four red marbles is:
35 / 495 = 7 / 99
B. More of the marbles are green than red:
In this case, we need to consider situations where either 3 or 4 marbles are green.
The probability of choosing 3 green marbles can be calculated as:
(C(5,3) * C(7,1)) / C(12,4)
C(5,3) = 5! / (3!(5-3)!) = 5! / (3!*2!) = (5*4) / (2*1) = 10
C(7,1) = 7! / (1!(7-1)!) = 7! / (1!*6!) = 7
So, the probability of choosing 3 green marbles is:
(10 * 7) / 495 = 70 / 495 = 14 / 99
The probability of choosing all 4 green marbles is:
C(5,4) / C(12,4) = (5! / (4!(5-4)!)) / (12! / (4!(12-4)!)) = (5! / (4!*1!)) / (12! / (4!*8!)) = (5*4) / (12*11*10*9) = 20 / 9,240 = 1 / 462
Therefore, the probability of choosing more green marbles than red is:
(14 / 99) + (1 / 462) = 53 / 462
C. Both red and green marbles are present:
Here, we need to consider two cases: 3 red and 1 green, or 2 red and 2 green marbles.
The probability of choosing 3 red and 1 green marbles is:
(C(7,3) * C(5,1)) / C(12,4) = (35 * 5) / 495 = 175 / 495 = 35 / 99
The probability of choosing 2 red and 2 green marbles is:
(C(7,2) * C(5,2)) / C(12,4) = (21 * 10) / 495 = 210 / 495 = 14 / 33
Therefore, the probability of both red and green marbles being present is:
(35 / 99) + (14 / 33) = 105 / 297 + 98 / 297 = 203 / 297
D. Two of the marbles chosen are both labeled "5":
There are 2 cases to consider: the first two marbles are both labeled "5," or the last two marbles are both labeled "5."
The probability of selecting a "5" marble on the first draw and another "5" marble on the second draw is:
(C(2,2) * C(10,2)) / C(12,4) = (1 * 45) / 495 = 45 / 495 = 1 / 11
The probability of selecting two non-"5" marbles on the first two draws and then the last two marbles being both labeled "5" is:
(C(10,2) * C(2,2)) / C(12,4) = (45 * 1) / 495 = 45 / 495 = 1 / 11
Therefore, the probability of two of the marbles chosen being both labeled "5" is:
(1 / 11) + (1 / 11) = 2 / 11
To find the probability of an event, we need to calculate the ratio of the number of desired outcomes to the number of possible outcomes.
A. To calculate the probability that all four marbles are red, we need to find the number of ways to select 4 red marbles out of the 7 available, divided by the total number of ways to select any 4 marbles out of the total 12 marbles.
The number of ways to choose 4 red marbles is C(7,4) = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = 35.
The total number of ways to choose 4 marbles out of 12 is C(12,4) = 12! / (4! * (12-4)!) = 12! / (4! * 8!) = 495.
Therefore, the probability that all four marbles are red is 35 / 495 = 7 / 99.
B. To calculate the probability that more of the marbles are green than red, we need to find the number of ways to select at least 3 green marbles out of the 5 available and at most 1 red marble out of the 7 available, divided by the total number of ways to select any 4 marbles out of the total 12 marbles.
The number of ways to choose at least 3 green marbles out of 5 is C(5,3) + C(5,4) + C(5,5) = 10 + 5 + 1 = 16.
The number of ways to choose at most 1 red marble out of 7 is C(7,0) + C(7,1) = 1 + 7 = 8.
The total number of ways to choose 4 marbles out of 12 is 495 (as calculated in part A).
Therefore, the probability that more of the marbles are green than red is (16 * 8) / 495 = 128 / 495.
C. To calculate the probability that both red and green marbles are present, we need to find the number of ways to select at least 1 red marble out of the 7 available and at least 1 green marble out of the 5 available, divided by the total number of ways to select any 4 marbles out of the total 12 marbles.
The number of ways to choose at least 1 red marble out of 7 is 2^7 - 1 = 127 (since we can choose any number of red marbles from 1 to 7, excluding the case of choosing none).
The number of ways to choose at least 1 green marble out of 5 is 2^5 - 1 = 31 (since we can choose any number of green marbles from 1 to 5, excluding the case of choosing none).
The total number of ways to choose 4 marbles out of 12 is 495 (as calculated in part A).
Therefore, the probability that both red and green marbles are present is (127 * 31) / 495 = 3987 / 495 = 131 / 165.
D. To calculate the probability that two of the marbles chosen are both labeled "5", we need to find the number of ways to choose 2 "5" marbles out of 2 available, multiplied by the number of ways to choose any 2 marbles out of the remaining 10 marbles, divided by the total number of ways to select any 4 marbles out of the total 12 marbles.
The number of ways to choose 2 "5" marbles out of 2 available is C(2,2) = 1.
The number of ways to choose any 2 marbles out of the remaining 10 marbles is C(10,2) = 10! / (2! * (10-2)!) = 10! / (2! * 8!) = 45.
The total number of ways to choose 4 marbles out of 12 is 495 (as calculated in part A).
Therefore, the probability that two of the marbles chosen are both labeled "5" is (1 * 45) / 495 = 9 / 99 = 1 / 11.