A bag contains 7 red marbles, 6 white marbles, and 10 blue marbles. You draw 4 marbles out at random, without replacement. What is the probability that all the marbles are red?

What is the probability that exactly two of the marbles are red?

What is the probability that none of the marbles are red?

To calculate the probability of drawing 4 red marbles out of the bag without replacement, we can use the concept of combinations.

Total number of marbles in the bag = 7 red + 6 white + 10 blue = 23 marbles

1) Probability of drawing all red marbles:
Number of ways to choose 4 red marbles = C(7, 4) = 7! / (4! * (7-4)!) = 35
Number of ways to choose 4 marbles out of 23 = C(23, 4) = 23! / (4! * (23-4)!) = 8855

Therefore, the probability of drawing all red marbles = 35 / 8855 ≈ 0.00395 or 0.395%.

2) Probability of drawing exactly two red marbles:
First, we need to calculate the number of ways to choose 2 red marbles from 7 red marbles = C(7, 2) = 7! / (2! * (7-2)!) = 21
Next, the number of ways to choose 2 marbles out of the remaining 16 marbles (6 white + 10 blue) = C(16, 2) = 16! / (2! * (16-2)!) = 120

Hence, the probability of drawing exactly two red marbles = (21 * 120) / 8855 ≈ 0.28477 or 28.477%.

3) Probability of drawing no red marbles (all marbles are not red):
Number of ways to choose 4 marbles out of the remaining 16 marbles (6 white + 10 blue) = C(16, 4) = 16! / (4! * (16-4)!) = 1820
Total number of ways to choose 4 marbles out of 23 = C(23, 4) = 8855

Therefore, the probability of drawing no red marbles = 1820 / 8855 ≈ 0.20543 or 20.543%.

To find the probability of different outcomes when drawing marbles from a bag without replacement, we need to use the concept of combinations.

The total number of marbles in the bag is 7 + 6 + 10 = 23.

1. Probability of drawing all red marbles:
The number of ways to draw 4 red marbles out of 7 is given by the combination formula: C(7, 4) = 7! / (4! * (7-4)!) = 35.
The total number of ways to draw any 4 marbles out of 23 is given by the combination formula: C(23, 4) = 23! / (4! * (23-4)!) = 8855.
Therefore, the probability of drawing all red marbles is 35/8855 ≈ 0.00395.

2. Probability of exactly two red marbles:
The number of ways to choose 2 red marbles out of 7 is given by C(7, 2) = 7! / (2! * (7-2)!) = 21.
The number of ways to choose 2 other marbles out of 16 (the remaining marbles after removing the 2 red ones) is given by C(16, 2) = 16! / (2! * (16-2)!) = 120.
The total number of ways to choose any 4 marbles out of 23 is still 8855.
Therefore, the probability of exactly two red marbles is (21 * 120) / 8855 ≈ 0.283.

3. Probability of none of the marbles being red:
The number of ways to choose 4 marbles out of the 16 non-red ones is given by C(16, 4) = 16! / (4! * (16-4)!) = 1820.
The total number of ways to choose any 4 marbles out of 23 is still 8855.
Therefore, the probability of none of the marbles being red is 1820 / 8855 ≈ 0.2059.

Please note that these calculations assume that all marbles have an equal chance of being drawn and that the drawing is done randomly.

prob(probability that all the marbles are red)

= (7/23)(6/22)(5/21)(4/20) = 1/253

or, by combinations:
prob(all 4 red) = C(7,4) / C(23,4) = 35/8855 = 1/253

b) probability that exactly two of the marbles are red
= C(7,2)*C(16,2) / C(23,4) = 21*120/8855 = 72/253

c) probability that none of the marbles are red ---> implies they are either white or blue
= C(16,4) / C(23,4) = ...