# Odds of drawing colored marbles from a bag

#### MonGoose

##### New member
Assume I have 60 marbles in a bag distributed equally among 5 colors. So 12 marbles per color.

I draw 5 marbles simultaneously (meaning I don't put any marbles back in the bag before drawing the next one).

• What are the odds of drawing exactly 1 red marble?
• What are the odds of drawing at least 2 red marbles?
I guess about 70% and 5%.

• How many marbles do I need to draw simultaneously to have at least 84% chance of drawing at least 1 red marble?
• Assuming the answer is X, what are the odds of drawing at least 2 red marbles when drawing X marbles simultaneously from the bag?
Thanks for your help.

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#### JeffM

##### Elite Member
If you do not tell us how you got your answers, how can we guess what you are doing wrong?

It might help if you started by looking at

#### MonGoose

##### New member
If you do not tell us how you got your answers, how can we guess what you are doing wrong?

It might help if you started by looking at

I assumed I could get a solution / calculation here. I assumed wrong.

#### JeffM

##### Elite Member
I assumed I could get a solution / calculation here. I assumed wrong.
Yes, you did.

• Jomo

#### MonGoose

##### New member
Either that, or you just don't know the answer :-D

Here's a shot:

• What are the odds of drawing exactly 1 red marble?
=(12/60 * 48/59 * 47/58 * 46/57 * 45/56) * 5 * 100% = 43%

Is that correct?

Odds of first marble to be red are 12/60.
Odds of 1st 2nd 3rd 4th or 5th marble to be red is the same times 5.
Odds of the other four marbles to be not red are 48/59 x 47/58 x 46/57 x 45/56.

• What are the odds of drawing at least 2 red marbles?
Exactly two red =(12/60 x 11/59 x 48/58 x 47/57 x 46/56) x 10 x 100% = 21%
Exactly three red =(12/60 x 11/59 x 10/58 x 48/57 x 47/56) x 10 x 100% = 5%
Ignoring drawing 4-5 red, odds of drawing 2 or 3 red is 21 + 5 = 26%

• How many marbles do I need to draw simultaneously to have at least 84% chance of drawing at least 1 red marble?
8 marbles

 5 colors 8 marbles draw exactly 1 red 35%​ draw exactly 2 red +32%​ draw exactly 3 red +14%​ draw exactly 4 red +3%​ draw 1 to 4 red =84%​

• Assuming the answer is X, what are the odds of drawing at least 2 red marbles when drawing X marbles simultaneously from the bag?
32 + 14 + 3 + ... = about 50%

#### Dr.Peterson

##### Elite Member
Either that, or you just don't know the answer :-D

Here's a shot:

• What are the odds of drawing exactly 1 red marble?
=(12/60 * 48/59 * 47/58 * 46/57 * 45/56) * 5 * 100% = 43%

Is that correct?

Odds of first marble to be red are 12/60.
Odds of 1st 2nd 3rd 4th or 5th marble to be red is the same times 5.
Odds of the other four marbles to be not red are 48/59 x 47/58 x 46/57 x 45/56.
This is correct, except that what you are calculating is called probability, not odds. Those are two entirely different ways to present an answer.

• What are the odds of drawing at least 2 red marbles?
Exactly two red =(12/60 x 11/59 x 48/58 x 47/57 x 46/56) x 10 x 100% = 21%
Exactly three red =(12/60 x 11/59 x 10/58 x 48/57 x 47/56) x 10 x 100% = 5%
Ignoring drawing 4-5 red, odds of drawing 2 or 3 red is 21 + 5 = 26%
There is a better way: First find the probability of drawing no red marbles, or exactly 1 red marble, and subtract from 1. Do you see why?

How many marbles do I need to draw simultaneously to have at least 84% chance of drawing at least 1 red marble?
The better method is even better for this. You want the probability of drawing no red marbles to be at most 16%.

#### MonGoose

##### New member
Thanks. Glad I figured that out on my own.

Ignoring drawing 4-5 red, odds of drawing 2 or 3 red is 21 + 5 = 26%

There is a better way: First find the probability of drawing no red marbles, or exactly 1 red marble, and subtract from 1. Do you see why?
Yes. But how exactly?

Drawing no red marbles = (48/60 * 47/59 * 46/58 * 45/57 * 44/56) = 0,31 -> Is that correct?

Drawing exactly 1 red = (12/60 * 48/59 * 47/58 * 46/57 * 45/56) * 5 = 0,43

1-0.31-0.43 = 0.26 it does not equal 0.21., that's a difference of 5%

Would that mean drawing exactly 4 or 5 red = 5% ?

How many marbles do I need to draw simultaneously to have at least 84% chance of drawing at least 1 red marble?
The better method is even better for this. You want the probability of drawing no red marbles to be at most 16%.
Like this?

Not drawing red with 8 marbles = 48/60 * 47/59 * 46/58 * 45/57 * 44/56 * 43/55 * 42/54 * 41/53 * 100 = 15% ?

May I ask another thing: calculating the positions the marbles come out, is hard to keep track of once you start drawing 3+ red marbles.

RED RED RED other other
RED RED other RED other
RED RED other other RED
...
I mean the 10 in this example, which I counted on my fingers:
Exactly two red =(12/60 x 11/59 x 48/58 x 47/57 x 46/56) x 10 x 100% = 21%

How can you calculate these positions without resorting to manual work?

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#### Dr.Peterson

##### Elite Member
Yes. But how exactly?

Drawing no red marbles = (48/60 * 47/59 * 46/58 * 45/57 * 44/56) = 0,31 -> Is that correct?

Drawing exactly 1 red = (12/60 * 48/59 * 47/58 * 46/57 * 45/56) * 5 = 0,43

1-0.31-0.43 = 0.26 it does not equal 0.21., that's a difference of 5%

Would that mean drawing exactly 4 or 5 red = 5% ?
Probably. Calculate them and see ...

Like this?

Not drawing red with 8 marbles = 48/60 * 47/59 * 46/58 * 45/57 * 44/56 * 43/55 * 42/54 * 41/53 * 100 = 15% ?
Yes. So to find the 8, you can just keep multiplying one fraction after another until you get down below 16%.

May I ask another thing: calculating the positions the marbles come out, is hard to keep track of once you start drawing 3+ red marbles.

RED RED RED other other
RED RED other RED other
RED RED other other RED
...
I mean the 10 in this example, which I counted on my fingers:
Are you familiar with combinations? This is 5C3.

#### pka

##### Elite Member
I mean the 10 in this example, which I counted on my fingers:
How can you calculate these positions without resorting to manual work?
Position or order has absolutely nothing to do with this qustion.
This question is all about content. LOOK HERE.
That is the probability of drawing no red or exactly one red.
Subtract that number from one to get the probability at least two reds.