The probability that a randomly selected 2-digit number belongs to the set $\left\{n \in N:\left(2^{n}-2\right)\right.$ is a multiple of 3$\}$ is equal to :

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Probability

The probability that a randomly selected 2-digit number belongs to the set

{n \in N : (2^{n} - 2)

is a multiple of 3

}

is equal to :

Answer: (B)

Since

2 \equiv - 1 (mod 3)

2^{n} - 2 \equiv (- 1)^{n} + 1 (mod 3)

, which is

0

for odd

n

and

2

for even

n

. So the set consists of the odd natural numbers.
Of the

90

two-digit numbers

10, 11, \dots, 99

, exactly

45

are odd, so the probability is

\frac{45}{90} = \frac{1}{2}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option