My Math Forum Probability questions.

 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: 2,431 Thanks: 1315 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. $P[\text{at most 2 hands with no aces or kings}] = P[0]+P[1]+P[2] =\dbinom{28}{0} (1-p)^{28} +\dbinom{28}{1} p(1-p)^{27}+\dbinom{28}{2}p^2(1-p)^{26} \approx 0.595721$ 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. $P[\text{at most 2 hands with no aces or kings}] = P[0]+P[1]+P[2] =\dbinom{28}{0} (1-p)^{28} +\dbinom{28}{1} p(1-p)^{27}+\dbinom{28}{2}p^2(1-p)^{26} \approx 0.595721$ 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 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
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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 $P[0]$ $P[0] = \dfrac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda} \approx 0.111$
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

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