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 :
Solution
Total number of cases = ${}^{90}{C_1} = 90$<br><br>Now, ${2^n} - 2 = {(3 - 1)^n} - 2$<br><br>$${}^n{C_0}{3^n} - {}^n{C_1}{.3^{n - 1}} + .... + {( - 1)^{n - 1}}.{}^n{C_{n - 1}}3 + {( - 1)^n}.{}^n{C_n} - 2$$<br><br>= $$3\left( {{3^{n - 1}} - n{3^{n - 2}} + ... + {{( - 1)}^{n - 1}}.n} \right) + {( - 1)^n} - 2$$<br><br>$({2^n} - 2)$ is multiply of 3 only when n is odd<br><br>Required Probability $= {{45} \over {90}} = {1 \over 2}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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