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

Correct answer: (C) 1/2. Total two-digit numbers is 90 So, total number of cases =C190=90 Now, 2^n-2=3-1^n-2 =C0n3^n-C1n·3^(n-1)+…+-1^(n-1)·Cn-1n3+-1^n·Cnn-2 =33^(n-1)-n3^(n-2)+…+-1^(n-1)·n+-1^n-2 Here, the term -1^n-2 should be multiple of 3. So, 2^n-2 is multiple of 3 only when n is odd. There are total 45 odd two-digit numbers. Hence, the required probability =C145/90=45/90=1/2

Q48 / 90 · JEE Main 2021 (27 Jul Shift 1) · Mathematics

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

A
$\frac{1}{6}$
B
$\frac{2}{3}$
Correct answer
C
$\frac{1}{2}$
D
$\frac{1}{3}$

Solution

Total two-digit numbers is $90$

So, total number of cases $= {}^{90}C_{1} = 90$

Now, $2^{n} - 2 = {(3 - 1)}^{n} - 2$

$= {}^{n}C_{0} 3^{n} - {}^{n}C_{1} \cdot 3^{n - 1} + \dots + {(- 1)}^{n - 1} \cdot {}^{n}C_{n - 1} 3 + {(- 1)}^{n} \cdot {}^{n}C_{n} - 2$

$= 3 (3^{n - 1} - n 3^{n - 2} + \dots + {(- 1)}^{n - 1} \cdot n) + {(- 1)}^{n} - 2$

Here, the term ${(- 1)}^{n} - 2$ should be multiple of $3 .$

So, $(2^{n} - 2)$ is multiple of $3$ only when $n$ is odd.

There are total $45$ odd two-digit numbers.

Hence, the required probability $= \frac{{}^{45}C_{1}}{90} = \frac{45}{90} = \frac{1}{2}$

Hence, the correct answer is option: 'C'.

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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

Solution

Relevant Questions

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