Probability Question 383
Question: For the three events A, B and C, P (exactly one of the events A or B occurs)=P (exactly one of the two events B or C occurs)=P(exactly one of the events C or A occurs)=P and P (all the three events occur simultaneously) $ =P^{2} $ , where 0<p<1/2. Then the probability of at least one of the three events A, B and C occurring is
Options:
A) $ \frac{3p+2p^{2}}{2} $
B) $ \frac{p+3p^{2}}{4} $
C) $ \frac{p+3p^{2}}{2} $
D) $ \frac{3p+2p^{2}}{4} $
Show Answer
Answer:
Correct Answer: A
Solution:
We known that
P (exactly one of A or B occurs)
$ =P(A)+P(B)-2P(A\cap B) $
$ \therefore P(A)+P(B)-2P(A\cap B)=p $ - (1)
Similarly, $ P(B)+P(C)-2P(B\cap A)=p $ - (2)
and $ P(C)+P(A)-2P(C\cap A)=p $ - (3)
Adding Eqs. (1), (2), and (3) we get
$ 2[P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)]=3p $
$ \Rightarrow P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C) $
$ -P(C\cap A)]=3p/2 $ - (4)
It is also given that $ P(A\cap B\cap C)=p^{2} $ - (5)
Now,P(at least one of A, B and C)
$ =P(A)+p(B)+p(C)-p(A\cap B)-p(B\cap C) $
$ -P(C\cap B)+P(A\cap B\cap C) $
$ =\frac{3p}{2}+p^{2}=\frac{3p+2p^{2}}{2} $ [Using Eqs. (4) and (5)]
BETA
