Multivariable sample space

[Kariege]

Hi,
I'm currently having a lot of trouble with this probability problem. For example:
Suppose there are 5 balls in a bag with number 1,2,3,4,5. I pick a ball at random 20 times (with replacement).
Lets say the probability of each ball being picked is:
P(1) = 0.5
P(2) = 0.15
P(3) = 0.1
P(4) = 0.2
P(5) = 0.05
After I pick the ball 20 times, I sum it up. The sum is denoted as T.
I want to find the probability that T>=50. How do I go about doing this?

I'm not entirely sure if this actually links to sample space. Sample space is just something that I have in my mind. I've seen sample space where there are 2 variables, but what about more than 2?

Any help would be appreciated
Thanks

 

[mathman]

It looks messy. The distribution is multinomial. For each combination you need to calculate the sum and finally add up the probabilities for those where the sum is what you want.

Mutinomial term $\frac{n!}{i!j!k!l!m!}p_1^ip_2^jp_3^kp_4^lp_5^m$ over all non-negative possibilities where i+j+k+l+m=n. (n=20).

 

[statdad]

You could write a little routine to work out the actual sampling distribution for the sum (probably not the most efficient use of your time)
or use the Central Limit Theorem to get an approximation.
* You have the probability distribution for the number to be drawn on each selection - find the mean and variance
* The CLT says the sum is approximately normally distributed with mean = 20 times the mean found above and variance = 20 times the mean found above
Use the appropriate normal distribution for your approximation. (You may want to use a continuity correction: I haven't looked at any of the numbers)

 

[Kariege]

It looks messy. The distribution is multinomial. For each combination you need to calculate the sum and finally add up the probabilities for those where the sum is what you want.

Mutinomial term $\frac{n!}{i!j!k!l!m!}p_1^ip_2^jp_3^kp_4^lp_5^m$ over all non-negative possibilities where i+j+k+l+m=n. (n=20).

Thanks. In this context, how would you approach this using the formula?

Also why does i+j+k+l+m = 20? I'm confused.​

 

[statdad]

You could write a little routine to work out the actual sampling distribution for the sum (probably not the most efficient use of your time)
or use the Central Limit Theorem to get an approximation.
* You have the probability distribution for the number to be drawn on each selection - find the mean and variance
* The CLT says the sum is approximately normally distributed with mean = 20 times the mean found above and variance = 20 times the mean (<-- that should be variance: sorry) found above
Use the appropriate normal distribution for your approximation. (You may want to use a continuity correction: I haven't looked at any of the numbers)

 

[Kariege]

You could write a little routine to work out the actual sampling distribution for the sum (probably not the most efficient use of your time)
or use the Central Limit Theorem to get an approximation.
* You have the probability distribution for the number to be drawn on each selection - find the mean and variance
* The CLT says the sum is approximately normally distributed with mean = 20 times the mean found above and variance = 20 times the mean found above
Use the appropriate normal distribution for your approximation. (You may want to use a continuity correction: I haven't looked at any of the numbers)

Thanks for the reply.
CLT is quite a new concept to me but I think this can help me to solve the problem.
Sry if this is quite a stupid question but how would you find the mean and the variance in this case? Is it similar to finding the mean and variance of a frequency table because this is a probability table?

 

[statdad]

"Is it similar to finding the mean and variance of a frequency table because this is a probability table?"

Yes - make a table, row 1 the different numbers that could be picked, row 2 the probabilities you've assigned, and work as though they were frequencies. Here is an example (I simply picked the first page that popped up in a search)

http://www.mathsisfun.com/data/random-variables-mean-variance.html