JEE MAIN - Mathematics (2023 - 29th January Evening Shift - No. 6)
Let $$\mathrm{S} = \{ {w_1},{w_2},......\} $$ be the sample space associated to a random experiment. Let $$P({w_n}) = {{P({w_{n - 1}})} \over 2},n \ge 2$$. Let $$A = \{ 2k + 3l:k,l \in N\} $$ and $$B = \{ {w_n}:n \in A\} $$. Then P(B) is equal to :
$$\frac{3}{32}$$
$$\frac{1}{32}$$
$$\frac{1}{16}$$
$$\frac{3}{64}$$
Explanation
$$P({w_1}) + {{P({w_1})} \over 2} + {{P({w_1})} \over {{2^2}}}\, + \,..... = 1$$
$$\therefore$$ $$P({w_1}) = {1 \over 2}$$
Hence, $$P({w_n}) = {1 \over {{2^n}}}$$
Every number except 1, 2, 3, 4, 6 is representable in the form
$$2k + 3l$$ where $$k,l \in N$$.
$$\therefore$$ $$P(B) = 1 - P({w_1}) - P({w_2}) - P({w_3}) - P({w_4}) - P({w_6})$$
$$ = {3 \over {64}}$$
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