CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If P(E1)=0.2,P(E2)=0.4 and P(E3)=0.6 and E1,E2 and E3 are independent events, then the probability that at least one of these events E1,E2 and E3 occurs is

A
1.2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.048
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.808
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.952
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 0.808
P( at least one event E1,E2,E3 occurs )
=P(E1E2E3)=P(E1)+P(E2)+P(E3)P(E1E2)P(E2E3)P(E3E1)+P(E1E2E3)=P(E1)+P(E2)+P(E3)P(E1)P(E2)P(E2)P(E3)P(E3)P(E1)+P(E1)P(E2)P(E3)=0.2+0.4+0.6(0.2)(0.4)(0.4)(0.6)(0.6)(0.2)+(0.2)(0.4)(0.6)=0.808

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Probability
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon