Three numbers are chosen at random without replacement from {1,2,3,...,8}. The probability that their minimum is 3, given that their maximum is 6, is :
A
15
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
25
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
38
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A15 Let A be an event such that the maximum of the three randomly chosen numbers is 6
Let B be an event such that the minimum of the three randomly chosen numbers is 3 Probability of occurrence of eventA=P(A)=5C28C3 [The numbers<6are1,2,3,4,5] Probability of occurrence of eventB=P(B)=5C28C3 [The numbers>3are4,5,6,7,8]
The probability of choosing numbers which has minimum 3 and maximum 6 is P(A∩B)=2C18C3[∵3<4,5<6]
The probability that their minimum is 3, given that their maximum is 6 is P(BA)=P(A∩B)P(A)=2C15C2=210=15