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February 21st, 2018, 10:03 AM  #1 
Newbie Joined: Feb 2018 From: Canada Posts: 20 Thanks: 2  Probability questions.
Question 1: On a typical night at the bridge club on Tuesday evenings, we deal cards for 28 games (also called hands) of bridge. As each hand has thoroughly shuffled cards we can assume the cards are dealt independently for each hand. What is the probability that in 28 hands of bridge (wherein each hand I am dealt 13 cards), that I have at most 2 hands where I'm dealt no aces and no kings? This is my solution for this question, can someone take a look, thanks. Let X be the number of hands being dealt no Aces and no Kings. X follows Binomial Distribution with n=28 and p=(8C0)(44C13)/(52C13) = 0.08175498876. P(X<=2) = P(X=0) + P(X=1) + P(X=2) = (28C0)*(0.0818 )^0*(0.9182)^28 + (28C1)*(0.0818 )^1*(0.9182)^27 + (28C2)*(0.0818 )^2*(0.9182)^26 = 0.5957211. Question 2: Trees grow in a forest at a rate of 1.4 per 2 m^2. A man is standing in the forest. What is the probability the nearest tree is at least 1 meter away from him? 
February 21st, 2018, 07:28 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 923 
for each individual hand $p = P[\text{no aces or kings}] = \dfrac{\dbinom{44}{13}}{\dbinom{52}{13}} = \dfrac{15466}{189175} \approx 0.082$ The number of hands out of 28 that have no aces or kings has a binomial distribution with $n=28$ and $p$ as above. $P[\text{at most 2 hands with no aces or kings}] = P[0]+P[1]+P[2] =\dbinom{28}{0} (1p)^{28} +\dbinom{28}{1} p(1p)^{27}+\dbinom{28}{2}p^2(1p)^{26} \approx 0.595721$ Well done. 
February 21st, 2018, 09:58 PM  #3  
Newbie Joined: Feb 2018 From: Canada Posts: 20 Thanks: 2  Quote:
I think this one should follow A Poisson Distribution with a rate of 1.4/2 = 0.7. P(no tree within 1 m^2) = e^0.7 * 0.7^0 / 0! = 0.4966. Am I doing right? Thank you, you are very helpful.  
February 21st, 2018, 10:30 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 923  I've been thinking about it. I guess what we're supposed to do is note that a radial distance of $1~m$ corresponds to an area of $\pi ~m^2$ The expected number of trees in this area is $0.7\pi ~tree$ So we use a poisson distributed rv with $\lambda = 0.7\pi$ and compute $P[0]$ $P[0] = \dfrac{\lambda^0 e^{\lambda}}{0!} = e^{\lambda} \approx 0.111$ 
February 22nd, 2018, 06:40 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 923  Quote:
 
February 28th, 2018, 07:55 PM  #6 
Newbie Joined: Feb 2018 From: Canada Posts: 20 Thanks: 2  

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