Cumulative probability continued

  • Context: Undergrad 
  • Thread starter Thread starter Brad_U
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary
SUMMARY

This discussion focuses on calculating cumulative probability from multiple gumball machines with varying ratios of red and green gumballs. The example provided illustrates how to compute the expected number of gumballs in a bag after drawing one from each machine. Specifically, with two machines having 70% and 95% red gumballs, the expected values yield 1.65 green and 0.35 red gumballs, resulting in a probability of 0.825 for green and 0.175 for red when a gumball is drawn randomly from the bag. This method effectively demonstrates the average of the probabilities across multiple trials.

PREREQUISITES
  • Understanding of cumulative probability concepts
  • Familiarity with expectation values in probability theory
  • Basic knowledge of ratios and percentages
  • Ability to perform simple arithmetic calculations
NEXT STEPS
  • Research "Expectation value in probability theory" for deeper insights
  • Learn about "Cumulative distribution functions" for broader applications
  • Explore "Monte Carlo simulations" for practical probability experiments
  • Study "Bayesian probability" to understand advanced probability concepts
USEFUL FOR

Students in statistics, data analysts, and anyone interested in probability theory and its applications in real-world scenarios.

Brad_U
Messages
2
Reaction score
0
cumulative probability continued...

Ok. I have 25 gumball machines with red or green gumballs. Each machine has a different percentage of each. Number 1 has 70% red, 30% green. Number 2 has 95% red, 5% green, and so on. I buy 1 gumball from each an drop it into a bag. Then reach into the bag and pull a random gumball. How to calculate probability Red or Green from the bag?
 
Physics news on Phys.org


I would start calculating the expectation value of the green and red balls in the bag.

For example, if you draw one from each machine, you have on average (after repeating this experiment very very very often), (0,70 + 0,95 + ...) green balls in the bag, and (0,30 + 0,05 + ...) red ones.

Let's assume for this example that there are only two machines, so you will have 1,65 green ball and 0,35 red balls in a total of two.
If you draw one of the balls randomly, therefore, the probability of picking the green one is 1,65 / 2 = 0,825 and for the red one of course 0,175. (If you go back and check the calculation, this is basically the average of the probabilities, because we drew equal numbers of balls from each machine).
 


Thanks for the reply.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K