SOLUTION: A bag contains 8 red marbles, 9 white marbles, and 6 blue marbles. You draw 5 marbles out at random, without replacement. For the following questions, round the answer to 3 decimal

Question 1205258: A bag contains 8 red marbles, 9 white marbles, and 6 blue marbles. You draw 5 marbles out at random, without replacement. For the following questions, round the answer to 3 decimals as required. What is the probability that all the marbles are red?
The probability that all the marbles are red is
0.002
.
What is the probability that exactly two of the marbles are red?
The probability that exactly two of the marbles are red is
0.379
.
What is the probability that none of the marbles are red?
The probability of picking no red marbles is
0.089
.

are my answers correct? they're rounded to 3DP.
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!

The probability that all the marbles are red is rounds to 0.002 . What is the probability that exactly two of the marbles are red? The probability that exactly two of the marbles are red is rounds to 0.379 . What is the probability that none of the marbles are red? The probability of picking no red marbles is rounds to 0.089 . are my answers correct? they're rounded to 3DP. Yup! Edwin

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part 1
8 red + 9 white + 6 blue = 23 marbles total
A = P(1st is red) = 8/23
B = P(2nd is red given 1st is red) = 7/22
C = P(3rd is red given 1st 2 are red) = 6/21
D = P(4th is red given 1st 3 are red) = 5/20
E = P(5th is red given 1st 4 are red) = 4/19
Note the countdown of numerators and denominators.
I wouldn't reduce any of the fractions (or else you'll lose this countdown).
P(all 5 red) = A*B*C*D*E
P(all 5 red) = (8/23)*(7/22)*(6/21)*(5/20)*(4/19)
P(all 5 red) = 0.00166 approximately
P(all 5 red) = 0.002 when rounding to 3 decimal places
You are correct.

Part 2
8 red, 23 total, 23-8 = 15 marbles that aren't red
A = P(1st is red) = 8/23
B = P(2nd is red given 1st is red) = 7/22
C = P(3rd is not red given events A and B) = 15/21
D = P(4th is not red given previous events) = 14/20
E = P(5th is not red given previous events) = 13/19
P(1st two marbles are red) = A*B*C*D*E
P(1st two marbles are red) = (8/23)*(7/22)*(15/21)*(14/20)*(13/19)
P(1st two marbles are red) = 0.03786 approximately
That's just for the first two marbles being red.
But we have 5C2 = 10 combinations of arranging these red marbles.
Refer to the nCr combination formula.
We multiply that previous result by 10 to get 10*0.03786 = 0.3786 and that rounds to 0.379
You are correct.
This has a binomial-like feel to it, but it's not entirely the binomial probability formula (note how the trials are not independent).

Part 3
8 red, 23 total, 23-8 = 15 not red
A = P(1st is not red) = 15/23
B = P(2nd is not red given event A happened) = 14/22
C = P(3rd is not red given events A,B happened) = 13/21
D = P(4th is not red given events A,B,C happened) = 12/20
E = P(5th is not red given events A,B,C,D happened) = 11/19
P(no red) = A*B*C*D*E
P(no red) = (15/23)*(14/22)*(13/21)*(12/20)*(11/19)
P(no red) = 0.08924 approximately
P(no red) = 0.089
You are correct.

Nice work on getting all three parts correct.

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