Assume that we have 20 students connect to one router in the room, and we know t
ID: 3755151 • Letter: A
Question
Assume that we have 20 students connect to one router in the room, and we know that the probability of one student who is downloading files is 0.3, and the probability of one student who is playing games is 0.2.
a) What’s the probability of having exact 5 students who are downloading files at the same time?
b) What’s the probability of having exact 5 students who are downloading files and another 5 students who are playing games at the same time?
c) What’s the probability of having more than 5 students who are downloading files at the same time?
Explanation / Answer
Here there are 20 students
Probability of one student downloading files = 0.3
Probability of one student not downloading files = 1-0.3=0.7
Probability of one student playing games = 0.2
Probability of one student not playing games =1- 0.2=0.8
a)The probability of having exact 5 students who are downloading files at the same time
From binomial distribution probability of r out of n ways:
P(r out of n) = n!/r!*(n-r)! * pr *(1-p)(n-r)
Here n=20
r=5
p=0.3
q=0.7
Sustituting the values
n!/r!*(n-r)! * pr *(1-p)(n-r)
=> (20!/5!15!)*(0.3)5 *(0.7)15
=>(20*19*17*16*15! /5!*15!)*(0.00243)*(0.0047)
=>(1860480/120)*(0.000011)
=>(15504*0.000011)
=>0.17
The probability of having exact 5 students who are downloading files at the same time=0.17
b)The probability of having exact 5 students who are downloading files at the same time
From binomial distribution probability of r out of n ways:
P(r out of n) = n!/r!*(n-r)! * pr *(1-p)(n-r)
Here n=20
r=5
p=0.3
q=0.7
Sustituting the values
n!/r!*(n-r)! * pr *(1-p)(n-r)
=> (20!/5!15!)*(0.3)5 *(0.7)15
=>(20*19*17*16*15! /5!*15!)*(0.00243)*(0.0047)
=>(1860480/120)*(0.000011)
=>(15504*0.000011)
=>0.17
Now the probability of
the probability of having exact 5 students who are downloading files and another 5 students who are playing games at the same time is
0.17*(Probability of 5 students playing games from remaining students)
Probability of 5 students playing games from remaining students=
n=15,r=5,p=0.2,q=0.8
=>n!/r!*(n-r)! * pr *(1-p)(n-r)
=>(15! /5!*10!)*(0.2)5 *(0.8)10
=>(3003)*(0.00032)*(0.10)
=>(0.1031)
Probability of 5 students playing games from remaining students=0.1031
c)The probability of having more than 5 students who are downloading files at the same time
P(x>5)=1-P(x<=5)
P(x<=5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)
Probability of one student downloading files = 0.3
Probability of one student not downloading files = 1-0.3=0.7
P(x=0)=(20!/0!20!)*(0.3)0 *(0.7)20=0.00079
P(x=1)=(20!/1!19!)*(0.3)1 *(0.7)19=20*0.3*0.719=0.0068
P(x=2)=(20!/2!18!)*(0.3)2 *(0.7)18 =(190*0.027*0.718)=0.027
P(x=3)=(20!/3!17!)*(0.3)3 *(0.7)17=(1140*0.0027*0.0021)=0.071
P(x=4)=(20!/4!16!)*(0.3)4 *(0.7)16 =(4845*0.0081*0.0032)=0.12
P(x=5)=(20!/5!15!)*(0.3)5 *(0.7)15 =(15504*0.000011)=0.17
P(x<=5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)=0.00079+0.0068+0.027+0.071+0.12+0.17=0.395
P(x>5)=1-P(x<=5)=1-0.395=0.604
The probability of having more than 5 students who are downloading files at the same time P(x>5)=0.604