Question

please help me understand this problem.

k people in a room. We would like to find the probability that any two people 5. Suppose that there are have birthdays within a day of each other (which we'll call "near-day birthdays"). The number of pairs and the probability that any two people are born within a day of each other is of people would be S4Therefore, we would expect the average number of near-day birthday pairs to be be a Poisson random variable that represents the number of near-day birthday pairs with mean A 3 Let Y 365 3 365 Suppose that there are 20 people in a room. What is the probability that no two people have near-day birthdays? (Note: If you are using EXCEL, you may find the COMBIN command useful). b. Use your answer in part a) to find the probability that there is at least one pair of people who have near-day birthdays c. How many people would have to be gathered for there to be a greater than 50% chance that there is at least one pair of near-day birthdays? a.

Answer 1

Solutian Back up Theory:- Jf a sandom lariable Xnpoisson Ca)ie, X has poisson distribution hen with meor macs finctionCpme Sof 'x' is given by p(x =x e-. ). .... «) where x =0,, 2,.- ousing excel function, also be obtained This Payugayand Staisticel, POLSSON-.- l] CAn Nos to wokout the solution:- and hence EC 20c, Jx 33|36s Caiven Here K20 5616. Sa y poissan .s616)... ). Part la), ho too have Ryugwqaad PLY near-dag birtnday Cd and Prsbabi lity atleasi .to o have nean. dan binlindag dan iiday Ryzuqwqaug1 =1-6.9092 [vide angwu ho tuoo have ne an =0-1902 cs Scannedwith CamScanner

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