PES 2020 Crack With License Code Free Download ((BETTER))
公開日:2022/06/16 / 最終更新日:2022/06/16
PES 2020 Crack With License Code Free Download ((BETTER))
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PES 2020 Crack With License Code Free Download
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All images and screen shots are used from the original sources.Q:
Equality and Inequality in context of games
This is a general question. Some days back someone commented here that in order to prove the theorems in logic we use P(A and B) = P(A) * P(B). This fact made me think about the following examples:
In Tallying Game:
The probability that the number of tosses done till a ‘thriller’ is 4 is $1/4$ (as $ 4 = 2^4$). So the chances of a ‘thriller’ is $1/2 $. Hence the probability of a ‘thriller’ is 1/2. And the probability that the winning number is odd is 1/2.
In The winning number is odd:
This is a simple one as here $P(A) = P(A \cap B) + P(A \cap B^c)$, so the probability of $A$ is 1/2. But here is the problem. How do I get $P(A)$ by observing a single trial.
Question:
How do I find out the probability that a lucky toss will be the thriller in a Thallying Game? Should I take the probability of a thriller (which is 1/2), and add it to the probability that a lucky toss will be the winning number (which is 1/2) or vice-versa? This problem is similar to the question in the last part of the post.
A:
Hint: If the probability of the event $A$ is $1/2$, how can you find the probability of $A\cap B$?
Edit: For the second question, P(A) is just P(winning number is odd), and P(B) is P(lucky toss is thiller). So $$P(A\cap B)=P(winning number is odd)+P(lucky toss is thriller).$$
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