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

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In a binomial distribution game involving a bag of colored balls, players draw balls over 10 trials, earning $0.50 for each blue ball drawn. The goal is to determine the least value of k such that the probability of all 10 players advancing, having earned more than k dollars, is less than 0.1. The random variable X, representing the number of blue balls drawn, follows a binomial distribution X~B(10, 1/3). Calculations show that the smallest number of blue balls needed for the probability P(X ≤ n) to exceed 0.206 is n=2, leading to the conclusion that k must be set accordingly. The analysis concludes that k must be adjusted to ensure that the advancement probability criterion is met.
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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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