# Binomial distribution

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• April 14th 2012, 04:33 AM
Punch
Binomial distribution
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 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 $n=\frac{k}{0.50}$ $1-P(X$ $n) <0.794$ $P(X$ $n) > 0.206$ • April 14th 2012, 10:24 PM Punch Re: Binomial distribution Sorry! The missing part is: For each blue ball obtained, the player earns$0.50
• April 15th 2012, 12:11 AM
biffboy
Re: Binomial distribution
P(X=0)=0.01734 P(X=1)=0.0867 P(X=2)=0.1951 So P(X<or=2)= 0.01734+0.0867+0.1951=0.299

So n=2 and k=1