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

[jayshah19949596]

[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]

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?

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

 

[jayshah19949596]

b).

 

[pka]

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

b).

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]

I dont know about the third answer

 

[pka]

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 $\displaystyle \mathcal{P}(A\Rightarrow B)=\mathcal{P}(A^c\cup B)=\mathcal{P}(A^c)+\mathcal{P}(B)-\mathcal{P}(A^c)\cdot\mathcal{P}(B)$

 

[jayshah19949596]

What is A superscript C in

 

[pka]

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 $\displaystyle \mathcal{P}(A\Rightarrow B)=\mathcal{P}(A^c\cup B)=\mathcal{P}(A^c)+\mathcal{P}(B)-\mathcal{P}(A^c)\cdot\mathcal{P}(B)$

 

[jayshah19949596]

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

Is this The answer >????

 

[pka]

= 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]

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 A^c​ ?

 

[Steven G]

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

 

[pka]

A^c, 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]

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