Quantitative Aptitude MCQ - Probability
S = [0, 1, 2, ....,9)
Suppose we want a particular integer "3" to be
chosen.
9
Probability of not choosing "3" is 9/10 (in one
chance)
Probability of not choosing 3 in all the three
chances
Suppose we want a particular integer "3" to be
chosen.
9
Probability of not choosing "3" is 9/10 (in one
chance)
Probability of not choosing 3 in all the three
chances
Hence probability of choosing it in at least one
of the three chances
The five digits can be arranged in 5! ways, out
of which 4! will begin with zero.
of which 4! will begin with zero.
Total number of 5-figure numbers formed
= 5! - 4! = 96.
Those numbers formed will be divisible by 4
which will have two extreme right digits
divisible by 4,
i.e., numbers ending in 04, 12,20,24,32,40.
Now, numbers ending in 04 = 3! = 6,
numbers ending in 12 = 3! - 2! = 4,
numbers ending in 20 = 3! = 6,
numbers ending in 24 = 3! - 2! = 4,
numbers ending in 32 = 3! - 2! = 4,
and numbers ending in 40 = 3! = 6.
[Numbers having 12, 24, 32 in the extreme right
are (3! - 2!), since the numbers having zero on
t.he extreme left are to excluded.]
Total number of favourable ways
6 + 4 + 6 + 4 + 4 + 6 = 30
Hence, required probability = 
