
Ram and Shyam select two numbers from 1 to n. If the probability that Shyam selects a number is less than the number selected by Ram is $\dfrac{{63}}{{128}}$, then
A) n is odd
B) n is a perfect square
C) n is a cube
D) None of these
Answer
593.1k+ views
Hint:
The number of ways of selecting two numbers from n numbers is ${}^n{C_2}$.
So, the probability that probability that Shyam selects a number is less than the number selected by Ram is given by $\dfrac{{{}^n{C_2}}}{{n \times n}}$ .
Now, solve $\dfrac{{{}^n{C_2}}}{{n \times n}}$ and compare it with $\dfrac{{63}}{{128}}$.
On getting the value of n, look at the number if it is odd, perfect square or a cube.
Complete step by step solution:
It is given that Ram and Shyam select two numbers from 1 to n.
So, the number of ways of selecting two numbers from n numbers is ${}^n{C_2}$ .
Ram and Shyam both have the same number of choices i.e. n. So, the number of possibilities of Ram and Shyam selecting a number is $n \times n$ .
Now, the probability that probability that Shyam selects a number is less than the number selected by Ram is given by $\dfrac{{{}^n{C_2}}}{{n \times n}}$ .
$
\dfrac{{{}^n{C_2}}}{{n \times n}} = \dfrac{{\dfrac{{n!}}{{2!\left( {n - 2} \right)!}}}}{{{n^2}}} \\
= \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{2{n^2}\left( {n - 2} \right)!}} \\
= \dfrac{{n - 1}}{{2n}} \\
$
It is given that the probability that Shyam selects a number is less than the number selected by Ram is $\dfrac{{63}}{{128}}$ .
Thus, $\dfrac{{n - 1}}{{2n}} = \dfrac{{63}}{{128}}$
$
\therefore 128\left( {n - 1} \right) = 2 \times 63 \times n \\
\therefore 128n - 128 = 126n \\
\therefore 2n = 128 \\
\therefore n = 64 \\
$
Thus, we get n as 64, which is a square of 8.
So, option (B) is correct.
Note:
Here, the condition that Shyam selects a number is less than the number selected by Ram, does not affect the number of choosing two numbers from n numbers as we have directly counted the number of selecting two numbers from n numbers and then we assign the higher number to Ram and smaller number to Shyam.
The number of ways of selecting two numbers from n numbers is ${}^n{C_2}$.
So, the probability that probability that Shyam selects a number is less than the number selected by Ram is given by $\dfrac{{{}^n{C_2}}}{{n \times n}}$ .
Now, solve $\dfrac{{{}^n{C_2}}}{{n \times n}}$ and compare it with $\dfrac{{63}}{{128}}$.
On getting the value of n, look at the number if it is odd, perfect square or a cube.
Complete step by step solution:
It is given that Ram and Shyam select two numbers from 1 to n.
So, the number of ways of selecting two numbers from n numbers is ${}^n{C_2}$ .
Ram and Shyam both have the same number of choices i.e. n. So, the number of possibilities of Ram and Shyam selecting a number is $n \times n$ .
Now, the probability that probability that Shyam selects a number is less than the number selected by Ram is given by $\dfrac{{{}^n{C_2}}}{{n \times n}}$ .
$
\dfrac{{{}^n{C_2}}}{{n \times n}} = \dfrac{{\dfrac{{n!}}{{2!\left( {n - 2} \right)!}}}}{{{n^2}}} \\
= \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)!}}{{2{n^2}\left( {n - 2} \right)!}} \\
= \dfrac{{n - 1}}{{2n}} \\
$
It is given that the probability that Shyam selects a number is less than the number selected by Ram is $\dfrac{{63}}{{128}}$ .
Thus, $\dfrac{{n - 1}}{{2n}} = \dfrac{{63}}{{128}}$
$
\therefore 128\left( {n - 1} \right) = 2 \times 63 \times n \\
\therefore 128n - 128 = 126n \\
\therefore 2n = 128 \\
\therefore n = 64 \\
$
Thus, we get n as 64, which is a square of 8.
So, option (B) is correct.
Note:
Here, the condition that Shyam selects a number is less than the number selected by Ram, does not affect the number of choosing two numbers from n numbers as we have directly counted the number of selecting two numbers from n numbers and then we assign the higher number to Ram and smaller number to Shyam.
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