A bag contains 9 red marbles, 8 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. What is the probability that all the marbles are red?
1 The probability that all the marbles are red is...?
2 What is the probability that exactly two of the marbles are red?
The probability that exactly two of the marbles are red is..?
3 What is the probability that none of the marbles are red?
The probability of picking no red marbles is..?
^ that is actually wrong, because you are getting 4 marbles out of the bag...
I did the first 2 parts here:
http://www.jiskha.com/display.cgi?id=1447035932
#3.
prob(none of the 4 are red)
= C(14,4)/C(23,4)
= 1001/8855
= 13/115
To find the probability in each of these scenarios, we need to calculate the number of favorable outcomes (drawing the desired marbles) and divide it by the total number of possible outcomes.
1. The probability that all the marbles are red:
First, let's find the total number of marbles in the bag: 9 red + 8 white + 6 blue = 23 marbles.
Since we are drawing without replacement, the total number of ways to draw 4 marbles out of 23 is calculated as "23 choose 4" or C(23, 4) = (23! / (4! * (23-4)!)) = 8855.
The number of favorable outcomes (drawing all red marbles) is the number of ways to draw 4 marbles out of 9 red marbles, which is calculated as "9 choose 4" or C(9, 4) = (9! / (4! * (9-4)!)) = 126.
Therefore, the probability that all the marbles drawn are red is 126 / 8855 ≈ 0.0142 (approximately 1.42%).
2. The probability that exactly two of the marbles are red:
To find this probability, we need to calculate the total number of ways to draw 4 marbles out of 23 (as done before, C(23, 4) = 8855), and then find the number of favorable outcomes (drawing exactly two red marbles).
The number of favorable outcomes is calculated by multiplying the number of ways to choose 2 red marbles out of 9 (C(9, 2) = 36) by the number of ways to choose 2 non-red marbles out of the remaining 14 (C(14, 2) = 91).
Therefore, the number of favorable outcomes is 36 * 91 = 3276.
Hence, the probability that exactly two of the marbles drawn are red is 3276 / 8855 ≈ 0.3697 (approximately 36.97%).
3. The probability that none of the marbles are red:
Following the same steps as before, the number of favorable outcomes would be the number of ways to draw 4 marbles out of the 14 non-red marbles since no red marbles are allowed.
Thus, the number of favorable outcomes is calculated as "14 choose 4" or C(14, 4) = (14! / (4! * (14-4)!)) = 1001.
Therefore, the probability of not picking any red marbles is 1001 / 8855 ≈ 0.1131 (approximately 11.31%).