Question

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

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (A)

Total 2-digits number = 90

Total number of cases = ${}^{90}{C}_1=90$

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

${}^n{C}_0{3}^n-{}^{n-1}{C}_1{3}^{n-1}+........{(-1)}^{n-1}.{}^n{C}_{n-1}3+{(-1)}^n.{}^n{C}_n-2$

$\Rightarrow 3[3^{n-1}-n{3}^{n-2}+........{(-1)}^{n-1}.n]+{(-1)}^n-2$

$2^n-2$ is a multiple of 3 only when n is odd.

So, out of 90 2-digits number

half of them are odd number (i.e. 45)

So, probability $=\frac{45}{90}$

$=\frac{1}{2}$