The Probability of x Being Greater than 1 and Less than 2 Given a Specific Function f(n) a*n^2

In the field of probability theory, understanding how certain variables behave within specified intervals is crucial for various applications. This article will explore how to determine the probability of a variable ( x ) being greater than 1 and less than 2, given the function ( f(n) a*n^2 ). This exploration will involve several key steps, including function definition, calculation of specific values, normalization, and final probability determination.

Understanding the Function

The given function is ( f(n) a*n^2 ), where ( n ) can take on specific discrete values.

When ( n 0 ), ( f(0) a*0^2 0 ) When ( n 1 ), ( f(1) a*1^2 a ) When ( n 2 ), ( f(2) a*2^2 4a )

These values define the function for specific discrete values of ( n ), indicating that the function is defined for ( n 0, 1, 2 ).

Calculating the Function Values

To proceed, let's consider the function with the specific values provided:

1. When ( n 0 ), ( f(0) 0 )

2. When ( n 1 ), ( f(1) b )

3. When ( n 2 ), ( f(2) 4b )

Given that ( b ) is a constant, we need to determine the value of ( b ).

Understanding the Probability

To find the probability of ( x ) being in the interval ( (1, 2) ), we need to know the distribution of ( x ). If ( x ) takes values from ( {0, 1, 2} ) with probabilities defined by ( f(x) ), these probabilities must be normalized to ensure they sum to 1.

Normalization

The total probability must sum to 1:

( P(0) P(1) P(2) 1 )

Where:

( P(0) f(0) 0 )

( P(1) f(1) b )

( P(2) f(2) 4b )

This gives us:

( 0 b 4b 5b 1 )

Thus, ( b frac{1}{5} ).

Finding Probabilities

With the value of ( b ), we can now find the probabilities:

( P(1) b frac{1}{5} )

( P(2) 4b 4 * frac{1}{5} frac{4}{5} )

Calculating the Probability

Since ( x ) can only take the discrete values 0, 1, and 2, the probability of ( x ) being in the interval ( (1, 2) ), which includes only the value 2, is:

( P(1

Conclusion: The probability of ( x ) being greater than 1 and less than 2 is ( frac{4}{5} ).

Additionally, if the problem is modified to:

1. Let ( f(n) a*n^2 ) with ( a 1 ) and ( b 0 ).

2. Let the probability ( P(x 2) P(z 0) 0 ), then ( P(2) 1 ).

In this case, for the function to be a complete probability distribution function, all probabilities must add to one. Here, the function is defined as:

( f(0) a*0^2 0 )

It means ( n ) is never 0, so ( f(0) 0 ). Therefore, the function is defined only for ( n 1, 2 ).

Thus, the updated probability values are:

( P(1) b 0 )

( P(2) 1 )

Final Probability Calculation:

Since ( x ) can only be 1 or 2, and the probability of ( x ) being 2 is 1, the probability of ( x ) being in the interval ( (1, 2) ) is 0.

Conclusion: The probability of ( x ) being greater than 1 and less than 2 is 0.

Therefore, the correct probability in the modified context is: