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

# Four fair dice D1, D2, D3 and D4, each having six faces numbered 1,2,3,4,5 and 6, are rolled simultaneously. The probability that D4 shows a number appearing on one of D1, D2 and D3 is

A
91216
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
108216
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
125216
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
127216
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 91216Four fair dices rolled simultaneously.Total number of outcomes=6×6×6×6Suppose, the out come of the last one does not appear in the first three.The number of such outcomes possible=6×53(∵D1 can show any of 6 numbers and first three can show any of the remaining 5 numbers)Hence, probability that D4 does not show a number shown by any of D1,D2 or D3=6×5364So, the required probability=1−6×5364=91216

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Introduction
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program