Statistics And Probability Question 38
Question: The probability of the simultaneous occurrence of two events A and B is p. if the probability that exactly one of the events occurs is q, then which of the following is not correct?
Options:
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. $
Show Answer
Answer:
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. $
BETA
