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

For three events A, B and C, if P (exactly one of A or B occurs) = P(exactly one of B or C occurs) = P (exactly one of C or A occurs) =14 and P (all the three events occur simultaneosuly) =116, then the probability that atleast one of the events occurs is ?

A
732
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
716
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
764
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
316
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 716
We have, P(exactly one of A or B occurs)
=P(AB)P(AB)
=P(A)+P(B)2P(AB)
According to the question,
P(A)+P(B)2P(AB)=14 . . . (i)
P(B)+P(C)2P(BC)=14 . . . (ii)
and P(C)+P(A)2P(CA)=14 . . . (iii)
On adding Eqs. (i),(ii) and (iii), we get
2[P(A)+P(B)+P(C)P(AB)P(BC) P(CA)]=34
P(A)+P(B)+P(C)P(AB)P(BC) P(CA)=38
P(atleast one event occurs)
=P(ABC)=P(A)+P(B)+P(C)P(AB)P(BC) P(CA)+P(ABC)=38+116=716 (P(ABC)=116)

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