#### jayshah19949596

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- P(A and B).
- P(A or B).
- P(A => B).

Note that (A => B) is a Boolean expression. You are asked to compute the probability that (A => B) evaluates to true.

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- P(A and B).
- P(A or B).
- P(A => B).

Note that (A => B) is a Boolean expression. You are asked to compute the probability that (A => B) evaluates to true.

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Do you know what independence implies? Lets find out.A and B are two Boolean events that are independent of each other. P(A) = 0.3 and P(B) = 0.6.Compute the following three quantities:

- P(A and B).
- P(A or B).
- P(A => B).

Note that (A => B) is a Boolean expression. You are asked to compute the probability that (A => B) evaluates to true.

If \(\displaystyle H~\&~G\) are independent then what goes in the blank?

\(\displaystyle \Large{\mathcal{P}(H \cap G) = \mathcal{P}(H)\_\_\_\_\mathcal{P}(G),\;\quad a) + ,\;\quad b)\; \cdot ,\;\quad c) \cap ,\;\quad d \;\cup }\)

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b).

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Correct! \(\displaystyle \mathcal{P}(A \cap B) = \mathcal{P}(A)\cdot\mathcal{P}(B)\)

\(\displaystyle \begin{align*}\mathcal{P}(A \cup B)&=\mathcal{P}(A)+\mathcal{P}(B)-\mathcal{P}(A\cap B)\\&=\mathcal{P}(A)+\mathcal{P}(B)-\mathcal{P}(A)\cdot\mathcal{P}(B) \end{align*}\)

Now you have the first two.

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I dont know about the third answer

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In order to find \(\displaystyle \mathcal{P}(A\Rightarrow B)\), you must know basic logic.I dont know about the third answer

\(\displaystyle A \Rightarrow B\) is read in a logic course as "If A then B."

Then you need to \(\displaystyle (A\Rightarrow B)\equiv (A^c\vee B)~.\)

Therefore you need to \(\displaystyle \mathcal{P}(A\Rightarrow B)=\mathcal{P}(A^c\cup B)=\mathcal{P}(A^c)+\mathcal{P}(B)-\mathcal{P}(A^c)\cdot\mathcal{P}(B)\)

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What is A superscript C in

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I gave you the answer to the third one. Look at it again.I do not know the answer of the third one

In order to find \(\displaystyle \mathcal{P}(A\Rightarrow B)\), you must know basic logic.

\(\displaystyle A \Rightarrow B\) is read in a logic course as "If A then B."

Then you need to \(\displaystyle (A\Rightarrow B)\equiv (A^c\vee B)~.\)

Therefore you need to \(\displaystyle \mathcal{P}(A\Rightarrow B)=\mathcal{P}(A^c\cup B)=\mathcal{P}(A^c)+\mathcal{P}(B)-\mathcal{P}(A^c)\cdot\mathcal{P}(B)\)

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= 0.3 + 0.6 – (0.3) * (0.6)

= 0.9 - 0.18

= 0.72

Is this The answer >????

= 0.9 - 0.18

= 0.72

Is this The answer >????

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= 0.3 + 0.6 – (0.3) * (0.6)

= 0.9 - 0.18

= 0.72Is this The answer>????

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Whoops. It looks like the end of your question got cut off.What is A superscript C in

Are you asking for the meaning of symbol A

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There many other notations for the complement of A such as: \(\displaystyle \neg A,~\overline{A}\)A^{c}, A' and ~A all mean not A which is the same as the complement of A.

I have a large collection of basic probability textbooks. Each of the five notations are use in at least one in that collection. However, the most popular seems to be these: \(\displaystyle A',~\overline{A}~\&~A^c\).

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\(\displaystyle A^C\) is theWhat is A superscript C in