probability help!

natacha

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The probability that event A occurs is P(A) = 0.3. The event B is independent of A and P(B)=0.4.
Event C is defined to be the event that neither A or B occurs.

Calculate P(C
b5ba02a830e3da1c6c0fc7bd9b07bef4.png
A'), where A' is the event that A does not occur.
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The answer at the back of my book is 0.6 but i dont know how to get it so could someone show me how to start?
 
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Hello, natacha!

This is a Conditional Probability problem.
Are you familiar with Bayes' Theorem? .\(\displaystyle P(X\,|\,Y) \:=\:\dfrac{P(X \cap Y)}{P(Y)}\)


\(\displaystyle \text{The probability that event }A\text{ occurs is: } P(A) = 0.3\)
\(\displaystyle \text{The event }B\text{ is independent of }A\text{ and: } P(B) = 0.4\)
\(\displaystyle \text{Event }C\text{ is defined to be the event that neither }A\text{ nor }B\text{ occurs. }\)

\(\displaystyle \text{Calculate }P(C|A'),\text{ where }A'\text{ is the event that }A\text{ does not occur.}\)

\(\displaystyle \text{Answer: }\,0.6\)

We have: .\(\displaystyle \begin{Bmatrix}P(A) \:=\:0.3 && P(B) \:=\:0.4 \\ P(A') \:=\:0.7 && P(B') \:=\:0.6 \end{Bmatrix}\)

Also: .\(\displaystyle P(C) \:=\:p(A' \cap B') \:=\: (0.7)(0.6) \:=\:0.42\)


We want: .\(\displaystyle P(C|A') \:=\:\dfrac{P(C \cap A')}{P(A')} \) .[1]

The numerator is: .\(\displaystyle P(C \cap A') \:=\:p(C) \:=\:0.42\)

The denominator is: .\(\displaystyle P(A') \:=\:0.7\)

Therefore, [1] becomes: .\(\displaystyle P(C|A') \:=\:\dfrac{0.42}{0.7} \:=\;0.6\)
 
The probability that event A occurs is P(A) = 0.3. The event B is independent of A and P(B)=0.4.
Event C is defined to be the event that neither A or B occurs. Calculate P(C
b5ba02a830e3da1c6c0fc7bd9b07bef4.png
A'), where A' is the event that A does not occur.
The answer at the back of my book is 0.6

If \(\displaystyle A~\&~B\) are independent events then so are \(\displaystyle A'~\&~B'\).

Therefore \(\displaystyle \mathcal{P}(C)=\mathcal{P}(A')\cdot\mathcal{P}(B')\).

Can you finish?
 
If we are given A does not occur then to have C, we only need that B does not occur. Since we are told that the probability that B does occur is 0.4, the probability that B does not occur is 1- .4= .6.
 
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