1. ## venn

11.give that : P(A)=P(B)=P(C)=1/4,P(AB)=P(CB)=0 and P(AC)=1/8 ,compute the probability that at least one of the three events A,B or C will occur .
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12 . suppose that :P(A)=0.4 , P(B)=α ,P(AUB)= 0.7:
i.what is the value of α which makes A and B disjoint ?
ii. what is the value of αwhich makes A and B independent ?
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13 . show that:
i. P(ABC)=P(A|BC)P(B|C)P(C)

2. Originally Posted by compufatwa
11.give that : P(A)=P(B)=P(C)=1/4,P(AB)=P(CB)=0 and P(AC)=1/8 ,compute the probability that at least one of the three events A,B or C will occur .
Draw a diagram.

From the diagram it is clear that

p(A or B or C)= P(A)+P(C)- P(A and C) + P(B).

RonL

3. Hello, compufatwa!

Here's #12 . . .

12 . Given: . $P(A) =0.4,\;P(B)= \alpha,\;P(A \cup B) = 0.7$

a) What is the value of $\alpha$ which makes A and B disjoint?
We have: . $P(A \cup B) \;=\;P(A) + P(B) - P(A \cap B)$

Hence: . $0.7 \;=\;0.4 + \alpha - P(A \cap B)$ .[1]

If A and B are disjoint, then $P(A \cap B) = 0$

Then [1] becomes: . $0.7 \;=\;0.4 + \alpha - 0\quad\Rightarrow\quad\boxed{\alpha \,=\,0.3}$

b) What is the value of $\alpha$ which makes A and B independent ?

If A and B are independent, then: . $P(A \cap B) \:=\:P(A)\!\cdot\!P(B)$

Then [1] becomes: . $0.7 \;=\;0.4 + \alpha -P(A)\!\cdot\!P(B)\quad\Rightarrow\quad0.7 \;=\;0.4 + \alpha - (0.4)(\alpha)$

Therefore: . $0.3 \:=\:0.6\alpha\quad\Rightarrow\quad\boxed{\alpha \,=\,0.5}$

4. thanks lot Soroban and CaptainBlack for this info