wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For the three events A, B and C, P(at least one occurring)=34, P(at least two occurring)=12 and P(exactly two occurring)=25. Which of the following relations is (are) CORRECT ?

A
P(ABC)=110
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
P(AB)+P(BC)+P(CA)=75
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

P(A)+P(B)+P(C)=2720

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
P(A¯¯¯¯B¯¯¯¯C)+P(¯¯¯¯A¯¯¯¯BC)+P(¯¯¯¯AB¯¯¯¯C)=14
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D P(A¯¯¯¯B¯¯¯¯C)+P(¯¯¯¯A¯¯¯¯BC)+P(¯¯¯¯AB¯¯¯¯C)=14


Given P(ABC)=34 (1)
P(AB¯¯¯¯C)+P(A¯¯¯¯BC)+P(¯¯¯¯ABC)+P(ABC)=12 (2)
And P(AB¯¯¯¯C)+P(A¯¯¯¯BC)+P(¯¯¯¯ABC)=25 (3)
From (2)(3),
P(ABC)=1225=110P(ABC)=110 (4)

P(AB)+P(BC)+P(CA)
=P(AB¯¯¯¯C)+P(A¯¯¯¯BC)+P(¯¯¯¯ABC)+3P(ABC)
=25+3×110P(AB)+P(BC)+P(CA)=710 (5)

P(A)+P(B)+P(C)
=P(ABC)+P(AB)+P(BC)+P(CA)P(ABC)
From (1),(4) and (5),
P(A)+P(B)+P(C)=34+710110=2720

Also, P(A¯¯¯¯B¯¯¯¯C)+P(¯¯¯¯A¯¯¯¯BC)+P(¯¯¯¯AB¯¯¯¯C)
=P(A)+P(B)+P(C)2[P(AB)+P(BC)+P(CA)]+3P(ABC)
=27201410+310=520=14

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Venn Diagrams
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon