The probability of event A occurring is 10/10. The probability of event B occurring is 0/10. What is the entropy of this system?

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3 years ago

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1 answer:

IgorLugansk [536] · Answer 1 · 2023-01-17T23:31:26+03:00

The events A and B are the outcomes of the probabilities in the system

The entropy of the system is 0

<h3>How to determine the entropy?</h3>

The probabilities are given as:

P(A) = 10/10

P(B) = 0/10

The entropy is calculated using:

$H(x) = -\sum \limits^{n}_{i = 1} p_i * \log_2(p_i)$

So, we have

$H(x) = -10/10 * \log_2(10/10) - 0/10 * \log_2(0/10)$

Evaluate the products

$H(x) = -\log_2(10/10) - 0$

Evaluate the sum

$H(x) = -\log_2(10/10)$

Evaluate the quotient

$H(x) =- \log_2(1)$

Express 1 as 2^0

$H(x) =- \log_2(2^0)$

Apply the power rule of logarithm

$H(x) =- 0\log_2(2)$

Evaluate the product

H(x) =0

Hence, the entropy of the system is 0