Question 377 of 417EasyMCQ
The probability that a randomly selected 2-digit number belongs to the set {n \[\in\] N : (2n - 2) is a multiple of 3} is equal to :
Solutions · 1
Official solution
Total number of cases = 90C1=90
Now, 2n−2=(3−1)n−2
\[{}^n{C_0}{3^n} - {}^n{C_1}{.3^{n - 1}} + .... + {( - 1)^{n - 1}}.{}^n{C_{n - 1}}3 + {( - 1)^n}.{}^n{C_n} - 2\]
= \[3\left( {{3^{n - 1}} - n{3^{n - 2}} + ... + {{( - 1)}^{n - 1}}.n} \right) + {( - 1)^n} - 2\]
(2n−2) is multiply of 3 only when n is odd
Required Probability \[ = {{45} \over {90}} = {1 \over 2}\] Answer the question to unlock