Question

If Ram and Shyam select two numbers with replacement from the set 1 to n. If the probability that Shyam selects a number which is less than the number selected by Ram is \[\dfrac{63}{128}\], Then which of the following is true.

a) n is odd

b) n is a perfect square

c) n is a perfect $4^{th}$ power

d) None of these

Answer 1

Hint: Now to find probability we will find the total number of possible selections that Ram and Shyam can make. Now Ram and Shyam both have n choices to choose a number from 1 to n. Next we will need the number of cases in which Shyam selects a number which is less than the number selected by Ram. Once we have these two we can find the probability with the help of formula $\dfrac{\text{number of favorable outcome}}{\text{total number of outcomes}}$

Complete step by step answer:

Now we are given that Ram and Shyam select two numbers from the set 1 to n.

Now let us consider an event A such that Shyam selects a number which is less than the number selected by Ram. 

Now the total number between 1 to n is equal to n

Now we will select two numbers from 1 to n and assign the higher number to Ram and lower number to Shyam.

The number of ways of selecting two numbers from n numbers is $^{n}{{C}_{2}}$ 

Hence the number of ways in which A event happens is $\dfrac{n!}{(n-2)!2!}=\dfrac{n(n-1)(n-2)!}{(n-2)!2!}=\dfrac{n(n-1)}{2}$

Now n(A) = $\dfrac{n(n-1)}{2}$

Now there are n numbers between 1 to n. Hence, Ram has n choices and Shyam also has n choices to select a number. So the total number of possibility such that Ram and Shyam select a number from 1 to n is ${{n}^{2}}$

Now we have to find the probability such that Shyam selects a number which is less than the number selected by Ram. Hence we have to find the probability of event A.

Now we know that probability is $\dfrac{\text{number of favorable outcome}}{\text{total number of outcomes}}$

And here A is our favorable outcome and let S be the total outcome.

Hence we get $n(A)=\dfrac{n(n-1)}{2}$ and $n(S)={{n}^{2}}$

With the formula for probability we get \[P(A)=\dfrac{n(A)}{n(S)}=\dfrac{\dfrac{n(n-1)}{2}}{{{n}^{2}}}=\dfrac{n(n-1)}{2{{n}^{2}}}=\dfrac{(n-1)}{2n}\]

Now this probability is given to be \[\dfrac{63}{128}\]

Hence we have  $\dfrac{(n-1)}{2n}=\dfrac{63}{128}$

$ \Rightarrow \dfrac{(n-1)}{n}=\dfrac{63}{64} $

$ \Rightarrow 64(n-1)=63n $

$ \Rightarrow 64n-64=63n $ 

$ \Rightarrow n=64 $

Hence we get the value of n as 64. 

Now 64 is neither an odd number nor a perfect fourth power. But 64 is a perfect square since the square of 8 is 64.

So, the correct answer is “Option B”.

Note: While the condition that Shyam selects a number less than the number selected by Ram is given it doesn’t actually affect our calculation since we have directly counted the number of ways in which 2 numbers can be selected and then we assign the higher number to Ram and Lower number to Shyam.