A) \[P(A')+P(B')=2+2q-p\]
B) \[P(A')+P(B')=2-2p-q\]
C) \[P(A\cap B|A\cup B)=\frac{p}{p+q}\]
D) \[P(A'\cap B')=1-p-q.\]
Correct Answer: A
Solution :
| [a] It is given that |
| \[P(A\cap B)=p\] And \[P(A'\cap B)=+P(A\cap B')=q.\] |
| Therefore, since \[P(A'\cap B)=P(B)-P(A\cap B),\] |
| We get |
| \[q=P(B)-P(A\cap B)+P(A)-P(A\cap B)\] |
| \[\Rightarrow P(A)+P(B)=q+2p\] |
| \[\Rightarrow P(A')+P(B')=1-P(A)+1-P(B)\] |
| \[=2-q-2p,\] |
| Showing that [b] is correct. The answer [c] is also correct because |
| \[P(A\cap B|A\cup B)=\frac{P[(A\cap B)\cap (A\cup B)]}{P(A\cup B)}=\frac{P(A\cap B)}{P(A\cup B)}\]\[=\frac{P(A\cap B)}{P(A)+P(B)-P(A\cap B)}=\frac{p}{q+2p-p}=\frac{p}{p+q}\] |
| Finally, [d] is correct because |
| \[P(A'\cap B')=1-P(A\cup B)\] |
| \[=1-[P(A)+P(B)-P(A\cap B)]\] |
| \[=1-(q+2p-p)=1-p-q.\] |
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