Marbles in the bag help

[Tra_galloway]

You have a bag of 10 marbles of different colors. You draw a marble and record its color and replace it 50 times. w/ the following results 9 purple, 6 yellow, 16 orange and 19 green. How many of each is in the bag?

 

[tkhunny]

Insufficient information. The only thing you can say for sure is that there is at least one of purple, yellow, orange, and green.

There may be red.
There may be blue, black, or white.

You can speculate with the proportions of your sample, I suppose, but I would expect to see a good confidence interval with that.

 

[HallsofIvy]

You can, as TKHunny suggests, estimate the number of each kind of marble in the bag.

There are 10 marbles, total in the bag so if there were "m" purple marbles in the bag, you would expect to get an average of \(\displaystyle \frac{m}{10}(50)= 5m\) purple marbles in 50 tries. If 5m= 9 then m= 9/5. Of course, that's not the number of purple marbles since it is not an integer- it's only an estimate and an estimate lose to 2. Similarly, if there were "n" yellow marbles, in 50 trials we would expect about 5n yellow marbles. Since we got 6, we would expect n to be close to 6/5- and the integer closest to 6/5 is 1. We estimate that there is 1 yellow marble in the bag. Can you estimate the number of orange and green marbles in the bag?

Because these numbers must be integers, I'm not sure "confidence intervals" would really tell you much.

 

[JeffM]

You can, as TKHunny suggests, estimate the number of each kind of marble in the bag.

There are 10 marbles, total in the bag so if there were "m" purple marbles in the bag, you would expect to get an average of \(\displaystyle \frac{m}{10}(50)= 5m\) purple marbles in 50 tries. If 5m= 9 then m= 9/5. Of course, that's not the number of purple marbles since it is not an integer- it's only an estimate and an estimate lose to 2. Similarly, if there were "n" yellow marbles, in 50 trials we would expect about 5n yellow marbles. Since we got 6, we would expect n to be close to 6/5- and the integer closest to 6/5 is 1. We estimate that there is 1 yellow marble in the bag. Can you estimate the number of orange and green marbles in the bag?

Because these numbers must be integers, I'm not sure "confidence intervals" would really tell you much.

Halls

These are genuine questions rather than implied criticisms. I do not know enough statistics to make criticisms.

I have an intuition that the most reliance can be placed on the estimate of the green marbles so I would start there. Is that intuition valid?

If g of the 10 marbles were green, you would expect to get in 50 random draws a green marble:

\(\displaystyle 50 * \dfrac{g }{10} = 5g\ times \implies 19 \approx 5g \implies g = 4.\) That's your logic, and I follow it.

Could I not use Bayes Theorem to calculate the probability that the estimate of 4 is correct?

If we decide that 4 is the best estimate for the number of green marbles, should the remaining calculations be based on 6 not-green marbles and 31 not-green draws?

Genuinely curious.

 

[HallsofIvy]

Halls

These are genuine questions rather than implied criticisms. I do not know enough statistics to make criticisms.

I have an intuition that the most reliance can be placed on the estimate of the green marbles so I would start there. Is that intuition valid?

If g of the 10 marbles were green, you would expect to get in 50 random draws a green marble:

\(\displaystyle 50 * \dfrac{g }{10} = 5g\ times \implies 19 \approx 5g \implies g = 4.\) That's your logic, and I follow it.

Could I not use Bayes Theorem to calculate the probability that the estimate of 4 is correct?

You would, in fact, be doing the same thing again.

If we decide that 4 is the best estimate for the number of green marbles, should the remaining calculations be based on 6 not-green marbles and 31 not-green draws?

Genuinely curious.

I thought about that but because the numbers derived my way, 2 purple, 1 yellow, 3 orange, and 4 green add to 10, that is not necessary.

 

[JeffM]

You would, in fact, be doing the same thing again.

Thanks for the answers.