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

If A and B are independent events such that 0<P(A)<1,0<P(B)<1 then

A
A,B are mutually exclusive
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
A and ¯B are independent
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
¯A,¯B are independent
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
P(A/B)+P(¯A/B)=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
B A and ¯B are independent
C ¯A,¯B are independent
D P(A/B)+P(¯A/B)=1
If A and B are independent then P(AB)=P(A)P(B)nowP(A¯¯¯¯B)=P(A)P(AB)=P(A)P(A)P(B)=P(A){1P(B)}=P(A)P(¯¯¯¯B)
Hence A and ¯¯¯¯B are independent.
similarly P(¯¯¯¯A¯¯¯¯B)=1P(AB)=1P(A)P(B)+P(AB)=1P(A)P(B)+P(A)P(B)=1P(A)P(B){1P(A)}={1P(A)}{1P(B)}=P(¯¯¯¯A)P(¯¯¯¯B)
Hence ¯¯¯¯A and ¯¯¯¯B are independent.
and P(A/B)+P(¯¯¯¯A/B)=P(AB)P(B)+P(¯¯¯¯AB)P(B)=P(A)P(B)+P(¯¯¯¯A)P(B)P(B)=P(A)+P(¯¯¯¯A)=1
Option A is incorrect because P(AB)=P(A)P(B)0 since A/Q P(A)>0andP(B)>0

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