JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 15)
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 :
$${1 \over 6}$$
$${2 \over 3}$$
$${1 \over 2}$$
$${1 \over 3}$$
Explanation
Total number of cases = $${}^{90}{C_1} = 90$$
Now, $${2^n} - 2 = {(3 - 1)^n} - 2$$
$${}^n{C_0}{3^n} - {}^n{C_1}{.3^{n - 1}} + .... + {( - 1)^{n - 1}}.{}^n{C_{n - 1}}3 + {( - 1)^n}.{}^n{C_n} - 2$$
= $$3\left( {{3^{n - 1}} - n{3^{n - 2}} + ... + {{( - 1)}^{n - 1}}.n} \right) + {( - 1)^n} - 2$$
$$({2^n} - 2)$$ is multiply of 3 only when n is odd
Required Probability $$ = {{45} \over {90}} = {1 \over 2}$$
Now, $${2^n} - 2 = {(3 - 1)^n} - 2$$
$${}^n{C_0}{3^n} - {}^n{C_1}{.3^{n - 1}} + .... + {( - 1)^{n - 1}}.{}^n{C_{n - 1}}3 + {( - 1)^n}.{}^n{C_n} - 2$$
= $$3\left( {{3^{n - 1}} - n{3^{n - 2}} + ... + {{( - 1)}^{n - 1}}.n} \right) + {( - 1)^n} - 2$$
$$({2^n} - 2)$$ is multiply of 3 only when n is odd
Required Probability $$ = {{45} \over {90}} = {1 \over 2}$$
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