Question

Let S=$\left\{ 1,2,....,20 \right\}$. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is “nice” is:
(a) $\dfrac{6}{{{2}^{20}}}$
(b) $\dfrac{5}{{{2}^{20}}}$
(c) $\dfrac{4}{{{2}^{20}}}$
(d) $\dfrac{7}{{{2}^{20}}}$

Answer 1

Hint: First, before proceeding for this, we can see clearly that total number of ways of subset S to be formed from the 20 elements as ${{2}^{20}}$. Then, by using the sum of series for n elements is given by the formula $\dfrac{n\left( n+1 \right)}{2}$. Then, to get the probability, we must use the definition of probability which states that it is the ratio of the favourable outcomes to the total outcomes.

Complete step by step answer:
In this question, we are supposed to find the probability that a randomly chosen subset of S is “nice” when S=$\left\{ 1,2,....,20 \right\}$. A subset B of S is said to be “nice” and the sum of the elements of B is 203.
So, before proceeding for this, we can see clearly that total number of ways of subset S to be formed from the 20 elements as:
${{2}^{20}}$
Now, we are also given in the question with the condition that word “nice” comes if the sum of elements of B is 203.
Now, by using the sum of series for n elements is given by the formula:
$\dfrac{n\left( n+1 \right)}{2}$
Here, we have the value of n as 20, so we get the sum of the series as:
$\begin{align}
  & \dfrac{20\left( 20+1 \right)}{2} \\
 & \Rightarrow \dfrac{20\times 21}{2} \\
 & \Rightarrow 10\times 21 \\
 & \Rightarrow 210 \\
\end{align}$
So, to get the sum of series as 203 from the value obtained sum as 210, we get:
$210-7=203$
Now, we need to subtract the elements from the subset S whose sum is 7 as some of the examples are:
$\left\{ 7 \right\},\left\{ 1,6 \right\},\left\{ 2,5 \right\},\left\{ 3,4 \right\},\left\{ 1,2,4 \right\}$
So, we get the number of the samples whose sum equals to 7 as 5.
Now, to get the probability, we must use the definition of probability which states that it is the ratio of the favourable outcomes to the total outcomes.
So, from the above calculations, we get the favourable cases as 5 and total outcomes as ${{2}^{20}}$ which gives the probability that a randomly chosen subset of S is “nice” is:
$\dfrac{5}{{{2}^{20}}}$
Hence, option (b) is correct.

Note:
Now, to solve these type of the questions we need to be careful with the subset making from the condition in the question as sum is 7 which are taken as $\left\{ 7 \right\},\left\{ 1,6 \right\},\left\{ 2,5 \right\},\left\{ 3,4 \right\},\left\{ 1,2,4 \right\}$but if we consider $\left\{ 6,1 \right\}$also which increase the count of the number of cases. However, in the case of sets there is no cases for the arrangement either $\left\{ 6,1 \right\}$or $\left\{ 1,6 \right\}$will be considered as one case.