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

#### jayshah19949596

##### New member
[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

##### Elite Member
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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b).

#### pka

##### Elite Member
$$\displaystyle {\mathcal{P}(A \cap B) = \mathcal{P}(A)\_\_\_\_\mathcal{P}(B),\;\quad a) + ,\;\quad b)\; \cdot ,\;\quad c) \cap ,\;\quad d \;\cup }$$
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.

#### pka

##### Elite Member
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

##### New member
What is A superscript C in

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#### pka

##### Elite Member
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

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

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#### pka

##### Elite Member
= 0.3 + 0.6 – (0.3) * (0.6)
= 0.9 - 0.18
= 0.72 Is this The answer >????
No it is not! Try $$\displaystyle \Large{(0.7)+(0.6)-(0.7)(0.6)}$$

#### mmm4444bot

##### Super Moderator
Staff member
What is A superscript C in
Whoops. It looks like the end of your question got cut off.

Are you asking for the meaning of symbol Ac​ ?

#### Jomo

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

#### pka

##### Elite Member
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

##### Elite Member
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 $$\displaystyle P(A^C)= 1- p$$.