0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# If A and B are two events, the probability that exactly one of them occurs is given byP(A)+P(B)−2P(A∩B)P(A∩B)+2P(¯A∩B)P(A∪B)−P(A∩B)P(¯A)+P(¯B)−2P(¯A∩¯B)

A
P(A)+P(B)2P(AB)
B
P(AB)P(AB)
C
P(¯A)+P(¯B)2P(¯A¯B)
D
P(AB)+2P(¯AB)
Solution
Verified by Toppr

#### Exactly one of the events of EA and B is represented by A∩¯¯¯¯B+¯¯¯¯A∩Bhence P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)={P(A)−P(A∩B)}+{P(B)−P(A∩B)} =P(A)+P(B)−2P(A∩B)which is option (A).⇒P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=[P(A)+P(B)−P(A∩B)]−P(A∩B)=P(A∪B)−P(A∩B)which is option (C).P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=1−P(¯¯¯¯A)+1−P(¯¯¯¯B)−2{1−P(¯¯¯¯A∪¯¯¯¯B)}=2P(¯¯¯¯A∪¯¯¯¯B)−P(¯¯¯¯A)−P(¯¯¯¯B)={2P(¯¯¯¯A)+2P(¯¯¯¯B)−2P(¯¯¯¯A∩¯¯¯¯B)}−P(¯¯¯¯A)−P(¯¯¯¯B)=P(¯¯¯¯A)+P(¯¯¯¯B)−2P(¯¯¯¯A∩¯¯¯¯B)which is option (D).and P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=1−P(¯¯¯¯A∪B)+P(¯¯¯¯A∩B)∵¯¯¯¯¯¯¯¯¯¯¯¯¯A∩¯¯¯¯B=¯¯¯¯A∪Bby demorgan's law P(A∩¯¯¯¯B)=1−P(¯¯¯¯A∪B)⇒P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)={1−P(¯¯¯¯A)−P(B)+P(¯¯¯¯A∩B)}+P(¯¯¯¯A∩B)=P(A)−P(B)+2P(¯¯¯¯A∩B)

Was this answer helpful?
2
Similar Questions
Q1
If A and B are two events, the probability that at least one of them occurs is
(a) P(A) + P(B) – 2 P(A ∩ B)
(b) P(A) + P(B) – P(A ∩ B)
(c) P(A) + P(B) + P(A ∩ B)
(d) P(A) + P(B) + 2P(A ∩ B)
View Solution
Q2
Consider the following statements:
(1) P(¯AB)=P(¯A)+P(B)P(¯AB)
(2) P(A¯B)=P(¯B)P(AB)
(3) P(AB)=P(B)P(A|B)
Which of the above statements are correct?
View Solution
Q3
If A and B are two events such that P(AB)=0.65 and P(AB)=0.15, then P(¯A)+P(¯B)=
View Solution
Q4
Let A,B & C be 3 arbitary events defined on a sample space S and if, P(A)+P(B)+P(C)=p1,P(AB)+P(BC)+P(CA)=p2 & P(ABC)=p3, then the probability that exactly one of the three events occurs is given by:
View Solution
Q5
P(AB)=P(A)P(AB)=P(AB)P(B)=P(A¯B)=1P(¯AB)
View Solution