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 February 21st, 2018, 10:03 AM #1 Member   Joined: Feb 2018 From: Canada Posts: 46 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: 2,533 Thanks: 1390 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. $P[\text{at most 2 hands with no aces or kings}] = P+P+P =\dbinom{28}{0} (1-p)^{28} +\dbinom{28}{1} p(1-p)^{27}+\dbinom{28}{2}p^2(1-p)^{26} \approx 0.595721$ Well done. Thanks from Shanonhaliwell February 21st, 2018, 09:58 PM   #3
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 Originally Posted by romsek 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. $P[\text{at most 2 hands with no aces or kings}] = P+P+P =\dbinom{28}{0} (1-p)^{28} +\dbinom{28}{1} p(1-p)^{27}+\dbinom{28}{2}p^2(1-p)^{26} \approx 0.595721$ Well done.
Can you help me with the second question.
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
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 Originally Posted by Shanonhaliwell Can you help me with the second question?
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 $P$

$P = \dfrac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda} \approx 0.111$ February 22nd, 2018, 06:40 PM   #5
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 Originally Posted by romsek 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 $P$ $P = \dfrac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda} \approx 0.111$
I'd be interested in knowing if this is correct if you eventually find out. February 28th, 2018, 07:55 PM   #6
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 Originally Posted by romsek I'd be interested in knowing if this is correct if you eventually find out.
Sorry for the late reply. Your answer is correct. Thanks a million for your help. I don't know how to express my gratitute towards you for your help Tags probability, questions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fuyu1993 Advanced Statistics 1 November 4th, 2013 12:40 PM jabelone Advanced Statistics 3 April 27th, 2013 12:43 AM booster Probability and Statistics 4 November 26th, 2012 01:49 AM jhon13 Probability and Statistics 3 May 24th, 2012 01:35 AM Anonymous Coward Algebra 5 December 7th, 2009 01:13 PM

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