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

p₂ = 1 $$-$$ p (all three numbers are > 40)

$$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$$

So, $${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$$

p₁ = 1 $$-$$ p (all 3 numbers are $$\le$$ 80)

$$ = 1 - {\left( {{{80} \over {100}}} \right)^3} = {{125 - 64} \over {125}} = {{61} \over {125}}$$

So, $${{625{p_1}} \over 4} = {{625} \over 4} \times {{61} \over {125}} = {5 \over 4} \times 61 = 76.25$$

Let an event E of sum of outputs are perfect square (i.e., 4 or 9), so

E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}

and an event F of sum of outputs are prime numbers

(i.e., 2, 3, 5, 7, 11) so

F = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}

and event T of sum of outputs are odd numbers

(i.e., 3, 5, 7, 9, 11)

T = {(1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)}

Now, required probability $$p = P(T/E)$$

$$ = {{P(T \cap E)} \over {P(E)}}$$

where, $${P(T \cap E)}$$ = probability of occurring perfect square odd number before prime

$$ = \left( {{4 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{4 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{4 \over {36}}} \right) + ....\infty $$

$$ = {{{4 \over {36}}} \over {1 - {{14} \over {36}}}} = {4 \over {22}} = {2 \over {11}}$$

and P(E) = probability of occurring perfect square before prime

$$ = \left( {{7 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{7 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{7 \over {36}}} \right) + ....\infty $$

$$ = {{{7 \over {36}}} \over {1 - {{14} \over {36}}}} = {7 \over {22}}$$

$$ \therefore $$ $$P(T/E) = {{{2 \over {11}}} \over {{7 \over {22}}}} = {4 \over 7} = p \Rightarrow 14p = 8$$