QJEEAdv21Maf42d: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Maf42d Solution: A number of chosen at random from the set {1, 2, 3, .....,…

Given, set = {1, 2, 3, ...., 2000}

Let E₁ = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}

$\therefore$ n(E₁) = 666

and E₂ = Event that it is a multiple of 7 = {7, 14, ..., 1995}

$\therefore$ n(E₂) = 285

${E_1} \cap {E_2}$ = multiple of 21 = {21, 42, ....., 1995}

$n({E_1} \cap {E_2}) = 95$

$\therefore$ $P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$

$P({E_1} \cup {E_2}) = {{666 + 285 - 95} \over {2000}} = {{856} \over {2000}} = p$ (given)

Hence, $500p = 500 \times {{856} \over {2000}} = {{856} \over 4} = 214$