JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 15)
Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is ..........
Answer
8
Explanation
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
$$ = \left( {{7 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{7 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{7 \over {36}}} \right) + ....\infty $$
$$ = {{{7 \over {36}}} \over {1 - {{14} \over {36}}}} = {7 \over {22}}$$
$$ \therefore $$ $$P(T/E) = {{{2 \over {11}}} \over {{7 \over {22}}}} = {4 \over 7} = p \Rightarrow 14p = 8$$
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
$$ = \left( {{7 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{7 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{7 \over {36}}} \right) + ....\infty $$
$$ = {{{7 \over {36}}} \over {1 - {{14} \over {36}}}} = {7 \over {22}}$$
$$ \therefore $$ $$P(T/E) = {{{2 \over {11}}} \over {{7 \over {22}}}} = {4 \over 7} = p \Rightarrow 14p = 8$$
Comments (0)
