probability help!

[natacha]

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

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?

 

[soroban]

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) \:=\(A' \cap B') \:=\: (0.7)(0.6) \:=\:0.42\)


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

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

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

Therefore, [1] becomes: .$\displaystyle P(C|A') \:=\:\dfrac{0.42}{0.7} \:=\;0.6$

 

[pka]

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

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?

 

[HallsofIvy]

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.