A bag contains 6 red marbles, 9 white marbles, and 8 blue marbles. You draw 5 marbles out at random, without replacement.
Round your answers to 4 decimal places as needed.

1. What is the probability that all the marbles are red?

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

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

Problem 1

Work Shown:

There are n = 6 red marbles and r = 5 ways to pick them (order does not matter). Use the combination formula to find that

n C r = (n!)/(r!(n-r)!)

6 C 5 = (6!)/(5!*(6-5)!)

6 C 5 = (6!)/(5!*1!)

6 C 5 = (6*5!)/(5!*1!)

6 C 5 = (6)/(1!)

6 C 5 = (6)/(1)

6 C 5 = (6)/(1)

6 C 5 = 6

Now repeat for n = 6+9+8 = 23 and r = 5

n C r = (n!)/(r!(n-r)!)

23 C 5 = (23!)/(5!*(23-5)!)

23 C 5 = (23!)/(5!*18!)

23 C 5 = (23*22*21*20*19*18!)/(5!*18!)

23 C 5 = (23*22*21*20*19)/(5!)

23 C 5 = (23*22*21*20*19)/(5*4*3*2*1)

23 C 5 = (4037880)/(120)

23 C 5 = 33649

We have 6 ways of getting what we want (picking five red marbles) out of 33649 ways total (to get five marbles of any color). Again order does not matter.

Divide: 6/33649 = 0.00017831139112

This rounds to 0.0002 when rounding to four decimal places.

Problem 2

Work Shown:

There are 6 C 2 = 15 ways to pick 2 red marbles

There are 17 C 3 = 680 ways to pick 3 non-red marbles.

There are 15*680 = 10,200 ways to pick 5 marbles such that 2 are red, the rest aren't.

There are 33649 ways to select 5 marbles where color doesn't matter

So,

10200/33649 = 0.30312936491427

which rounds to 0.3031

Problem 3

Work Shown:

There are 17 marbles that aren't red, so there are 17 C 5 = 6188 ways to pull out five of them

This is out of 33649 ways to pull out five marbles in general.

6188/33649 = 0.18389848138131

which rounds to 0.1839

step-by-step explanation:

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