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?
What is the probability that exactly two of the marbles are red?
The probability that exactly two of the marbles are red is...?
9/23 bc there are 9 red marbles and 23 marbles altogether
^thats only for choosing one marble out of the bag
you have to solve for choosing 4 marbles out of the bag, and 2 of them are red
prob(4reds)
= (9/23)(8/22)(7/21)(6/20)
= 18/1265
or
prob(4reds)
= C(9,4) / C(23,4) = 126/8855 = 18/1265
= appr .014
prob(2 out of 4 are red)
= C(9,2) * C(14,2)/C(23,4)
= 36*91/8855
= 468/1265 = appr .37
To find the probability of drawing all red marbles, you first need to find the total number of ways to draw 4 marbles out of the 23 marbles in the bag (9 red, 8 white, and 6 blue). This can be calculated using the combination formula:
nCr = n! / (r!(n-r)!)
Where n is the total number of items and r is the number of items you want to choose.
So in this case, the number of ways to draw 4 marbles out of 23 is:
23C4 = 23! / (4!(23-4)!) = 23! / (4!19!) = (23 * 22 * 21 * 20) / (4 * 3 * 2 * 1) = 1771
Next, you need to calculate the number of ways to draw all red marbles. Since there are 9 red marbles in the bag, the number of ways to choose 4 red marbles is:
9C4 = 9! / (4!(9-4)!) = 9! / (4!5!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126
Therefore, the probability of drawing all red marbles is:
P(all red) = number of ways to draw all red marbles / number of ways to draw 4 marbles = 126 / 1771 ≈ 0.071
Now, let's calculate the probability of exactly two of the marbles being red. This can be done by finding the number of ways to choose 2 red marbles and 2 non-red marbles (white or blue), then dividing it by the total number of ways to draw 4 marbles.
To find the number of ways to choose exactly 2 red marbles, you can use the combination formula again. The number of ways to choose 2 red marbles out of 9 is:
9C2 = 9! / (2!(9-2)!) = 9! / (2!7!) = (9 * 8) / (2 * 1) = 36
Similarly, the number of ways to choose 2 non-red marbles out of 14 (8 white and 6 blue) is:
14C2 = 14! / (2!(14-2)!) = 14! / (2!12!) = (14 * 13) / (2 * 1) = 91
Therefore, the number of ways to choose exactly 2 red marbles and 2 non-red marbles is:
36 * 91 = 3276
Finally, the probability of exactly two of the marbles being red is:
P(exactly two red) = number of ways to choose exactly 2 red marbles and 2 non-red marbles / number of ways to draw 4 marbles = 3276 / 1771 ≈ 1.85