Statistics problem- please check my work

Post Reply New Topic

I am trying to work out the probability of something happening, just for the fun of it. I haven't had a math class since high school, which was several years ago. Can one of you math lovers tell me if I have this right? And if not, where did I mess up?

Let's say we have a big barrel of marbles. 10% of them are red, while the other 90% are blue. If you pull out two marbles at random, what is the probability that at least two of them will be red?

According to my calculations, the answer is 1 in 909.09... or .11% chance
I got that my multiplying 10 x 10 x 10 x 10 [the number of possible outcomes for each marble ^ the number of marbles drawn)
= 10,000 [the number of possible combinations of different types of marbles in groups of four]

I came up with 11 combinations of four marbles that would contain 2 or more red marbles:
rrbb
rbrb
rbbr
brrb
brbr
bbrr
rrrb
brrr
rbrr
rrbr
rrrr

So my answer is 11/10,000, which simplifies to 1/909.09. Is that right?

I assume you mean 4?

Sorry. Yes, I meant 4. I thought I had proofread that enough times.

Yourlist My list

rrbb rrrr
rbrb rrrb
rbbr rrbb
brrb rbbb
brbr bbbb
bbrr rrbr
rrrb rbrr
brrr rbrb
rbrr rbbr
rrbr brrr
rrrr brrb
brbr
brbb
bbrr
bbrb
bbbr
That is 2 options to the 4th power (number of balls)

I leave the remaining fun of the math to you!

Ciao!

My take: The probability of picking up one red ball is 0.1 (10%), and of getting one blue ball 0.9 (90%). You are correct in that there are 11 possible combinations:

rrbb
rbrb
rbbr
brrb
brbr
bbrr

rrrb
rrbr
rbrr
brrr

rrrr

The probability of getting the first option would be 0.1 x 0.1 x 0.9 x 0.9 = 0.0081. The next 5 options would give the same answer, therefore the probability of getting any of the first 6 options is 0.0081 x 6 = 0.0486

The probabilty of getting the 7th option, rrrb, would be 0.1 x 0.1 x 0.1 x 0.9 = 0.0009. The next 3 options would give the same answer, therefore the probability of getting any of these 4 options is 0.0009 x 4 = 0.0036

The probability of getting the final option is 0.1 x 0.1 x 0.1 x 0.1 = 0.0001.

Therefore the probabilty of getting any one of the 11 possibilities is 0.0486 + 0.0036 + 0.0001 = 0.0523.

I think this is incorrect. There aren't 10 possible outcomes for each marble, there are two: red and blue. The total number of possibilities is 16, not 10,000.




This step is also incorrect. You can only use 'Probability = (number of possibilities that satisfy our condition)/(total number of possibilities) if each individual possibility is equally likely, i.e. if rrbb is as likely as rrrb or rbbb.

As r and b are not equally likely you have to add up each probability like I did above.


Another technique that might speed things up (it's only a little faster in this problem, but can sometimes be much faster) is to first find out the opposite probability to what you want. Instead of finding the probability of at least 2 red balls, find the probability of at most 1 red balls. Our new list of possibilities is:

rbbb
brbb
bbrb
bbbr

bbbb

This is a shorter list than earlier, but we can find the probabilities the same way
The probability of the first one is 0.1 x 0.9 x 0.9 x 0.9 = 0.0729. Therefore the probability of the first 4 are 0.0729 x 4 = 0.2916
The probability of the final one is 0.9 x 0.9 x 0.9 x 0.9 = 0.6561
Therefore the total probability is 0.2916 + 0.6561 = 0.9477

If the probability of at most 1 red balls is 0.9477, then the probability of at least 2 red balls is 1 - 0.9477 = 0.0523

So here this method is only a little faster, but if the question instead asked "what is the probability of at least 1 red ball?" it would be much easier find the probability of no red balls first, and then take this probability away from 1.

Elligtonia, thank you for your explanation! That make a lot more sense than the answer I had.

Interesting thread. The problem needs to be stated more precisely, because as it stands, there are several answers that could be deemed "accurate."

Please specify if each marble is placed back in the back after being selected, or retained outside the bag. This naturally affects the probabilities in terms of independence and conditioning.

Further, specify the number of marbles, or at least if the number is sufficiently large for the purposes of the analysis here. In other words, are there enough marbles to pick four red ones.

The number of possible outcomes is of course 16 (i.e.: 2 to the fourth power). However, you don't specify if you stop picking marbles upon reaching two red ones. A better way to diagram this kind of probability analysis is using a branching tree diagram. It's also easier to do the analysis if you give precise counts to your marbles, rather than a ratio of 90% blue and 10% red.

Good luck with the analysis. Probability can be a lot of fun.

There's a slight problem with the premise of this problem. You've stated that we have a big barrel full of marbles, 10% red and 90% blue. So, when you pick the first marble, the chance that this marble is red is 10%, and blue 90%. But after that, its no longer an accurate statement to say that 10% of the marbles are red - you've just slightly changed the ratio of red/blue marbles remaining in the barrel by removing one. Without knowing the precise number of marbles in the barrel to start, the problem doesn't quite make sense. If you said that there's 1000 marbles to start, and you remove a red one, now you have 99 reds with 900 blues, so the proportion of reds is now slightly less than 10%.

edit: of course, that's not true if you put each marble back in the barrel after you've chosen it, which is probably what you're asking.

ScrewyWabbit is correct. If you only have 10 marbles in the box, then only one is red, and it is impossible to pull out a second red marble. If you have 20 marbles then there is a 10% chance to pull out the first one but less than 5% chance to pull out the second. If you have a million marbles then it is still very close to 10% for the second pull, and your odds would be around 12% by my estimation. But I'm not very good at math.

Last edited by Tsunami on 23 Apr 2013, 12:18 pm, edited 1 time in total.

Someone told me you do it this way. The chances of getting a combination with all reds is 0.10x0.10x0.10x0.10=0.0001, three red is 0.90x0.1x0.1x0.1=.0009, two reds is 0.9x0.9x0.1x0.1=0.0081, one red is 0.9x0.9x0.9x0.1=0.0729, and no reds is 0.9x0.9x0.9x0.9=0.6561. Now you have to add up all the successful combinations. There is 1 with all reds, 4 with three reds, and 6 with two reds, so 1x0.0001 + 4x0.0009 + 6x0.0081 = 5.23%

This doesn't take into account not replacing the balls, so the less balls are in your bucket the lower it goes, down to 0% (impossible) at 10/less than 20.