Question

Solve the following question and mark the best possible option.
Let x, y, z are 3 numbers (not necessarily distinct) chosen randomly with replacement. Also, x, y, z ∈ {1,2,3,4,5}. What is the probability that xy + z is even?

Options

• `2/5`

• `1/2`

• `64/125`

• `59/125`

Answer 1

2 cases arise:-
Case-1: xy & z both are odd
⇒ x, y, z all are odd
⇒ x, y, z = 1 or 3 or 5
So, probability = `3/5 xx 3/5 xx 3/5 = 27/125`

Case-2: Both xy & z are even.
Probability of z being even = `2/5`
Probability of xy being even = 1 - P(xy being odd)
= 1- `3/5 xx 3/5 = 16/25`.
So, required probability is `27/125 + 2/5 × 16/25= (27+32)/125 = 59/125`.