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

# Three number are chosen at random without replacement from {1,2,3,...,10}. The probability that minimum of the chosen number is 3 or their maximum is 7, cannot exceed

A
11/30
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
11/40
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1/7
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
1/8
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct options are A 11/30 C 11/40Let A and B denote the following events:A: minimum of the chosen number is 3.B: maximum of the chose number is 7.We have P(A)=P (choosing 3 and two other numbers from 4 to 10)=7C210C3=7×62×3×210×9×8=740P(B)=P (choosing 7 and two other numbers from 1 to 6)=6C210C3=6×52×3×210×9×8=18P(A∩B)=P (choosing 3 and 7 and one other number from 4 to 6).=310C3=3×3×210×9×8=140.Now, P(A∪B)=P(A)+P(B)−P(A∩B)=740+18−140=1140<1130.

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