The probability that a randomly selected 2 -digit number belongs to the set n ∈ N:(2n-2).. is a multiple of 3 is equal to

Question

The probability that a randomly selected 2 -digit number belongs to the set n ∈ N:(2n-2).. is a multiple of 3 is equal to (A) (1/6) (B) (2/3) (C) (1/2) (D) (1/3)

Tardigrade Admin · Accepted Answer

Total number of cases = 90 C1=90 Now, 2n-2=(3-1)n-2 n C0 3n- n C1 ⋅ 3n-1+ ldots .+(-1)n-1 ⋅ n Cn-1 3+(-1)n ⋅ n Cn-2 3(3n-1-n 3n-2+ ldots ldots+(-1)n-1 ⋅ n)+(-1)n-2 (2n-2) is multiply of 3 only when n is odd Req. Probability =(45/90)=(1/2)

Q. 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

$\frac{1}{6}$

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{3}$

Q. The probability that a randomly selected 2 -digit number belongs to the set {n∈N:(2n−2). is a multiple of 3} is equal to

61​

32​

21​

31​

Total number of cases =90C1​=90 Now, 2n−2=(3−1)n−2 nC0​3n−nC1​⋅3n−1+….+(−1)n−1⋅nCn−1​3+(−1)n⋅nCn​−2 3(3n−1−n3n−2+……+(−1)n−1⋅n)+(−1)n−2 (2n−2) is multiply of 3 only when n is odd Req. Probability =9045​=21​

Q. 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

$\frac{1}{6}$

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{3}$