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

AI Thread Summary
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
 
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