When the email was sent as a group email
Answer:
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
H(X)=5.4224 bits per symb
H(X|Y="not C")=0.54902 bits per symb
Explanation:
P(B)=2P(C)
P(A)=2P(B)
But
P(A)+P(B)+P(C)=1
4P(C)+2P(C)+P(C)=1
P(C)=1/7
Then
P(A)=4/7
P(B)=2/7
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
iii
If X={A,B,C}
and P(Xi)={4/7,2/7,1/7}
where Id =logarithm to base 2
Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}
=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}
=5.4224 bits per symb
if P(C) =0
P(A)=2P(B)
P(B)=1/3
P(A)=2/3
H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)
=0.54902 bits per symb
I don’t think we should because we all have our one choices that we should achieved and we should show
Answer:
Step by step explanation along with code and output is provided below
Explanation:
#include<iostream>
using namespace std;
// print_seconds function that takes three input arguments hours, mints, and seconds. There are 60*60=3600 seconds in one hour and 60 seconds in a minute. Total seconds will be addition of these three
void print_seconds(int hours, int mints, int seconds)
{
int total_seconds= hours*3600 + mints*60 + seconds;
cout<<"Total seconds are: "<<total_seconds<<endl;
}
// test code
// user inputs hours, minutes and seconds and can also leave any of them by entering 0 that will not effect the program. Then function print_seconds is called to calculate and print the total seconds.
int main()
{
int h,m,s;
cout<<"enter hours if any or enter 0"<<endl;
cin>>h;
cout<<"enter mints if any or enter 0"<<endl;
cin>>m;
cout<<"enter seconds if any or enter 0"<<endl;
cin>>s;
print_seconds(h,m,s);
return 0;
}
Output:
enter hours if any or enter 0
2
enter mints if any or enter 0
25
enter seconds if any or enter 0
10
Total seconds are: 8710