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Question: If Ram and Shyam select two numbers with replacement from the set 1 to n. If the probability that Sh...

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 63128\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 4th4^{th} power

d) None of these

Explanation

Solution

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 number of favorable outcometotal number of outcomes\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 nC2^{n}{{C}_{2}}

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

Now n(A) = n(n−1)2\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 n2{{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 number of favorable outcometotal number of outcomes\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)=n(n−1)2n(A)=\dfrac{n(n-1)}{2} and n(S)=n2n(S)={{n}^{2}}

With the formula for probability we get P(A)=n(A)n(S)=n(n−1)2n2=n(n−1)2n2=(n−1)2nP(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 63128\dfrac{63}{128}

Hence we have (n−1)2n=63128\dfrac{(n-1)}{2n}=\dfrac{63}{128}

⇒(n−1)n=6364\Rightarrow \dfrac{(n-1)}{n}=\dfrac{63}{64}

⇒64(n−1)=63n\Rightarrow 64(n-1)=63n

⇒64n−64=63n\Rightarrow 64n-64=63n

⇒n=64\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.