Quantitative Aptitude - Probability
Let A and B be two events defined as
Then P (A ∩ B) will be
i.e., P(E1 ∪ E2 ∪ E3 .... ∪ En)
= P(E1) + P(E2) +.... + P(En)
This is
- Option : B
- Explanation : 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
Hence probability of choosing it in at least one
of the three chances
- Option : B
- Explanation : The five digits can be arranged in 5! ways, out
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 =
