Probability: 2 Boolean events w/ P(A) = 0.3 and P(B) = 0.6

jayshah19949596

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[FONT=&quot]A and B are two Boolean events that are independent of each other. P(A) = 0.3 and P(B) = 0.6.[/FONT][FONT=&quot]Compute the following three quantities:[/FONT]

  • 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.
 

pka

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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.
Do you know what independence implies? Lets find out.
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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jayshah19949596

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b).
 

pka

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\(\displaystyle {\mathcal{P}(A \cap B) = \mathcal{P}(A)\_\_\_\_\mathcal{P}(B),\;\quad a) + ,\;\quad b)\; \cdot ,\;\quad c) \cap ,\;\quad d \;\cup }\)
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. You tell us about the third one. We await your trying.
 

jayshah19949596

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I dont know about the third answer
 

pka

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I dont know about the third answer
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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jayshah19949596

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What is A superscript C in
 
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pka

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I do not know the answer of the third one
I gave you the answer to the third one. Look at it again.
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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jayshah19949596

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= 0.3 + 0.6 – (0.3) * (0.6)
= 0.9 - 0.18
= 0.72

Is this The answer >????
 
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pka

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mmm4444bot

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Jomo

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Ac, A' and ~A all mean not A which is the same as the complement of A.
 

pka

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

HallsofIvy

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What is A superscript C in
\(\displaystyle A^C\) is the complement of event A- the event that A does not happen. If \(\displaystyle P(A)= p\) then \(\displaystyle P(A^C)= 1- p\).
 
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