there are two baskets, in the first, 4 red , 8 white balls, in the second 7 red and 6 white balls, two balls are taken from each baskets. the outcome of which of the two experiments, should be considered more uncertain

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Chapter1: Combinatorial Analysis
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there are two baskets, in the first, 4 red , 8 white balls, in the second 7 red and 6 white balls, two balls are taken from each baskets. the outcome of which of the two experiments, should be considered more uncertain?

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there are two baskets, in the first, 4 red , 8 white balls, in the second 7 red and 6 white balls, two balls are taken from each baskets. the outcome of which of the two experiments, should be considered more uncertain?

*can I the solution of this question same like the example in the picture

Example 4. There are two baskets. In the first - 5 red, 7 white balls. In the second -
8 red and 6 white balls. Two balls are taken from each baskets. The outcome of
which of the two experiments should be considered more uncertain?
Solution. Consider the first experiment, drawing two balls from the first urn, as a
system X with three outcomes:
исходы х
X1
X2
23
(2 кр.ш.) (1 кр.п.+1 бел.ш.) | (2 бел.ш.)
вероятности исходов | i i =iii+h n = wit=1
Найдем энтропию системы Х
132
= 11
n() + n() + n() 0,4118 +0,4854 +0,5256 =
H(X)
1, 4228 бит
=
132
Consider the second experiment, drawing two balls from the second basket, as a
system Y with three outcomes:
исходы уз
у
(2 кр.ш.) (1 кр.ш.+1 бел.ш.) (2 бел.ш.)
вероятности исходов
Let's find the entropy of the system Y:
H(Y) = n (i) + n (1) + 7 (9) = 0,5230 +0,4870 +0,4277
1, 4377 бит.
Because H(X) <H(Y), then the outcome of the second experiment is more
uncertain
Transcribed Image Text:Example 4. There are two baskets. In the first - 5 red, 7 white balls. In the second - 8 red and 6 white balls. Two balls are taken from each baskets. The outcome of which of the two experiments should be considered more uncertain? Solution. Consider the first experiment, drawing two balls from the first urn, as a system X with three outcomes: исходы х X1 X2 23 (2 кр.ш.) (1 кр.п.+1 бел.ш.) | (2 бел.ш.) вероятности исходов | i i =iii+h n = wit=1 Найдем энтропию системы Х 132 = 11 n() + n() + n() 0,4118 +0,4854 +0,5256 = H(X) 1, 4228 бит = 132 Consider the second experiment, drawing two balls from the second basket, as a system Y with three outcomes: исходы уз у (2 кр.ш.) (1 кр.ш.+1 бел.ш.) (2 бел.ш.) вероятности исходов Let's find the entropy of the system Y: H(Y) = n (i) + n (1) + 7 (9) = 0,5230 +0,4870 +0,4277 1, 4377 бит. Because H(X) <H(Y), then the outcome of the second experiment is more uncertain
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