00:02
In this question, solution here is entropy is equals to summation i is equals to 1 up to c minus of p i log of p i.
00:18
So, for the first part here h of past is equals to minus of p of t log 2 p of t minus p of f log 2 p of f.
00:37
So, here number of t is equals to 4 and number of f is equals to 2.
00:47
So, probability of t is equals to 4 divided by 6 and probability of f is equals to 2 divided by 6.
00:58
So, h of past is equals to minus of 4 divided by 6 log 2 4 divided by 6 minus 2 divided by 6 log 2 of 2 divided by 6 which is equals to 0 .3899 plus 0 .5283 that is equals to 0 .9182.
01:28
Therefore, h of past is equals to 0 .9182.
01:36
This is the answer to first part of this question.
01:39
Now, for the second part here h of past when gpa is given.
01:48
So, here h of past when gpa is equals to l is equals to minus of p of t when l is given log 2 probability of t when l is given minus probability of f when l is given log 2 f when l is given.
02:18
So, this is equals to minus of 1 divided by 2 log 2 of 1 divided by 2 minus 1 divided by 2 log 2 of 1 divided by 2 that is equals to 0 .5 plus 0 .5 which is equals to 1.
02:37
Therefore, h of past when gpa is equals to m is equals to minus of p of t when m is given log 2 probability of t when m is given minus probability of f when m is given log 2 multiplied by f of f when m is given.
03:11
So, this is equals to minus of 1 divided by 2 log 2 1 divided by 2 minus 1 divided by 2 log 2 of 1 divided by 2 that is equals to 0 .5 plus 0 .5 that is equals to 1.
03:29
Therefore, h of past when gpa is equals to h.
03:39
So, this is equals to minus of probability t when h is given log 2 t when h is given minus probability of f when h is given log 2 f when h is given...