A man walks into a factory, he knows that the factory contains 11 blue, yellow and red ball making machines, 10 of them are the standards which produce blue, yellow and red balls randomly with the probabilities in the ratio of 1:4:5 for blue yellow and red respectively. The other machine is special and at any time is equally likely to produce a ball of any of the colors. The man chooses one of the machines at random and turns it on. After a short while the man turns the machine off and looks at the balls it has produced. He sees that there are m balls of each color. Find the smallest value of m such that the man can know with at least 99% certainty that he was using the special machine.

2

wrong

well Im not too sure what the question is exactly asking for, but I assume you mean by 99% certainty that it is more than 99x more likely that it is the "special machine" rather than the "normal machine". If my assumption is correct, then the minimum value for m is 7

your assumption is correct, but your answer is incorrect.

Not quite sure if this is right, but m = 12?

Yes, m=12 is correct.

If anyone wants I can post the solution

I'll just post my method then:

Given that we have an equal number of balls m of each color from the machine, then let P_n(m) be the probability of this occurring and picking one of the normal machines, and P_s(m) be the analogous probability for the special machine.

Now let us consider the probability of each machine yielding a ball of each individual color. For the special machine, since each color has an equal chance of being given out, the probability of getting a ball of any color is 1/3, since there are three colors in total. Now consider the situation of getting m balls of each color. To find the probability of this happening, all we need to do is use the multiplicative law of probability for each ball.

That is, (1/3) ^ m * (1/3) ^ m * (1/3) ^ m

Which is (1/3 * 1/3 * 1/3) ^ m = (1/27) ^ m

We also need to take into account the number of machines. Since only 1 out of 11 of the machines are special, there is a 1/11 chance of picking the special machine at random.

Thus, we have P_s(m) = 1/11 * (1/27) ^ m

For the normal machine, we apply a similar principle. However, the individual probabilities are no longer equal. If we just apply the ratio given in the problem, then we have these probabilities:

blue = 1/10
yellow = 4/10 = 2/5
red = 5/10 = 1/2

Just like before, to find the probability of getting m balls of each color from the machine, we multiply the probabilities:

(1/10 * 2/5 * 1/2) ^ m = (1/50) ^ m

And since 10 out of the 11 machines are normal, P_n(m) = 10/11 * (1/50) ^ m

Finally, we need to determine when we are 99% certain that we picked from the special machine given our case of m balls. Using ultimifier's assumption, it is sufficient to find the smallest value of m such that:

P_s(m) >= 99 * P_n(m)

Or more simply,

P_s(m) / P_n(m) >= 99

Let's go ahead and simplify this using some algebra:

P_s(m) / P_n(m) = (1/11 * (1/27) ^ m) / (10/11 * (1/50) ^ m)
= [(1/11)/(10/11)] * [(1/27) ^ m / (1/50) ^ m]
= (1/10) * (50/27) ^ m

From here I simply used brute force (I wrote a program to check the value of this function starting at m = 1, stopping when it was greater than or equal to 99) and got the final result:

m = 12

aaah. i forgot to include the fact that there are 12 machines and the stuff past there. nice solution bass

yeah, that was pretty much my solution, though at the last part I avoided brute force by taking the log base 50/27 of 990 and rounding up.