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April 13th, 2018, 07:26 PM  #1 
Newbie Joined: Feb 2018 From: Canada Posts: 28 Thanks: 2  Probabaility question.
Today, I was having a short test and there were two problems that I was not able to figure them out. Question 1) In a certain population, the number of colds a person gets in a year has a Poisson distribution with mean 3 colds. A new anticold drug lowers $\lambda$ from 3 to 0.75 and is effective for 8 out of 10 people. At the beginning of last year, the entire population was given the drug. At the end of last year, one person was selected at random from the population and was found to have had only one cold during the year. What is the probability that the drug was effective for this person? Question 2) For the soccer team photo, the Freak Lunchboxes need to line up in two rows(6 players in the back row and 6 players in the front row). a) What is the probability Joe is in the front row? b) How many ways can this be done if Freddie and Erin need to be in a different row than Christine and Lauren? c) How many ways can this be done if Erin and Guillaume refuse to sit together? For part c of question 2, I was thinking that the answer = number ways they do not sit together in one row + number ways they sit in two rows = (number of ways in one row  number of ways they sit together) + (number ways they sit in two rows). 
April 13th, 2018, 07:59 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,981 Thanks: 1027 
1) $P[A  B]P[B] = P[BA]P[A]$ let $A$ be the drug is effective, let $B$ be the sample had 1 cold during the year. $P[BA] = poisson_{0.75}(1)$ $P[A] = 0.8$ $P[B] = poisson_{0.75}(1) + poisson_{3}(1)$ $P[AB] =\dfrac{(0.8)poisson_{0.75}(1)}{poisson_{0.75}(1) + poisson_{3}(1)} = 0.562747$ next problem in separate post 
April 13th, 2018, 08:02 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,981 Thanks: 1027  
April 13th, 2018, 08:28 PM  #4  
Newbie Joined: Feb 2018 From: Canada Posts: 28 Thanks: 2  Quote:
I think in this part, there are 2 cases: The first case is that they do not sit next to each other in one row. And the second case is that they sit in two rows which means that they do not sit together. By the way, for the first question I think $P[AB] =\dfrac{P(A\cap B)}{P(B)} = \dfrac{P(A)P(BA)}{P(A)P(BA)+P(A\complement )P(BA\complement )}$ $=\dfrac{0.8\dfrac{e^{0.75}0.75^1}{1!}}{0.8\dfrac{e^{0.75}0.75^1}{1!} + 0.2\dfrac{e^{3}3^1}{1!}}$ Last edited by Shanonhaliwell; April 13th, 2018 at 08:54 PM.  
April 13th, 2018, 08:51 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 1,981 Thanks: 1027 
ok, there's probably a clever way to do this but it's a small enough problem that brute force isn't too inefficient. If one of them sits in an edge seat, there are 10 seats available for the other. If one sits in an interior seat there are 9 seats available for the other. There are 4 edge seats and 8 interior seats. Thus there are 112 combinations available for assigning these two to two seats. There are 10 seats remaining in each of these combinations with 10! ways to seat the remaining 10 players. so there are $N = 112(10!) = 406425600$ ways to seat the players in problem (c) 
April 13th, 2018, 09:03 PM  #6  
Newbie Joined: Feb 2018 From: Canada Posts: 28 Thanks: 2  Quote:
So in one row, we fix one edge by having Erin or Guillaume sitting there. So now there are 10 people left, we choose 5 people $\binom{10}{5}$ then we will have $5!$ ways to arrange these 5 people.  
April 13th, 2018, 10:50 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 1,981 Thanks: 1027 
[QUOTE=romsek;592292] If one of them sits in an edge seat, there are 10 seats available for the other. If one sits in an interior seat there are 9 seats available for the other. There are 4 edge seats and 8 interior seats. Thus there are 112 combinations available for assigning these two to two seats. $4 \cdot 10 + 8 \cdot 9 = 40 + 72 = 112$ 
April 16th, 2018, 06:47 PM  #8  
Newbie Joined: Feb 2018 From: Canada Posts: 28 Thanks: 2 
[QUOTE=romsek;592303] Quote:
What a wonderful answer. Your solution is even better than my professsor's. My professor's answer is: Total of ways  ways together = $12!  2*\dbinom{10}{4}(5!)(2!)(6!)$ Last edited by Shanonhaliwell; April 16th, 2018 at 06:51 PM.  

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