What is the Least Value of K for Advancement in a Binomial Distribution Game?

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SUMMARY

The discussion focuses on determining the least value of k in a binomial distribution game involving a bag containing 4 red, 5 blue, and 6 green balls. Each player earns $0.50 for each blue ball drawn in 10 trials, modeled as X~B(10, 1/3). The condition for advancement is that the probability of all 10 players earning more than k dollars must be less than 0.1. The analysis reveals that the smallest number of blue balls drawn (n=2) satisfies the condition P(X ≤ n) > 0.206, leading to the conclusion that k must be set accordingly.

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  • Understanding of binomial distributions, specifically B(10, 1/3).
  • Knowledge of probability theory and cumulative distribution functions.
  • Familiarity with statistical concepts such as expected value and variance.
  • Ability to perform calculations involving probabilities and combinatorial analysis.
NEXT STEPS
  • Study the properties of binomial distributions, particularly B(n, p).
  • Learn how to calculate cumulative probabilities for binomial distributions.
  • Explore the concept of expected value in the context of random trials.
  • Investigate advanced probability techniques, such as the Central Limit Theorem.
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Mathematicians, statisticians, game theorists, and anyone involved in probability modeling and analysis in gaming or statistical experiments.

Punch
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A bag contains 4 red, 5 blue and 6 green balls. The balls are indistinguishable except for their colour. A trial consists of drawing a ball at random from the bag, noting its colour and replacing it in the bag. A game is plated by performing 10 trials in all.

At the start of the tournament, each player plays the above game once. Players who earned more than k dollars proceed to the next round. Find the least value of k such that, in a random sample of 10 players, the probability that all 10 players proceed to the next round is less than 0.1.

Let X be the number of blue balls drew.

X~B(10,$\frac{1}{3}$)

$[P(X>n)]^{10} < 0.1$ where $n=\frac{k}{0.50}$

$1-P(X $≤ $n) <0.794$

$P(X $≤ $n) > 0.206$
 
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Punch said:
A bag contains 4 red, 5 blue and 6 green balls. The balls are indistinguishable except for their colour. A trial consists of drawing a ball at random from the bag, noting its colour and replacing it in the bag. A game is plated by performing 10 trials in all.

At the start of the tournament, each player plays the above game once. Players who earned more than k dollars proceed to the next round. Find the least value of k such that, in a random sample of 10 players, the probability that all 10 players proceed to the next round is less than 0.1.

Let X be the number of blue balls drew.

X~B(10,$\frac{1}{3}$)

$[P(X>n)]^{10} < 0.1$ where $n=\frac{k}{0.50}$

$1-P(X $≤ $n) <0.794$

$P(X $≤ $n) > 0.206$

Incomplete question. Please include all the relevant information to the question in the thread with the question.

CB
 
CaptainBlack said:
Incomplete question. Please include all the relevant information to the question in the thread with the question.

CB

Sorry! The missing part is: For each blue ball obtained, the player earns $0.50
 
Punch said:
Sorry! The missing part is: For each blue ball obtained, the player earns $0.50

OK, so make a table of b(i,10,1/3):

Code:
            i     b(i,10,1/3)
            ----------------
            0     0.0173415 
            1     0.0867076 
            2      0.195092 
            3      0.260123 
            4      0.227608 
            5      0.136565 
            6     0.0569019 
            7     0.0162577 
            8    0.00304832 
            9   0.000338702 
           10  1.69351e-005

Now you need another column with the cumulative sum ...

(n=2 is the smallest number of wins such that P(X<=n)>0.206)

CB
 

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