Question

Two boys are asked to select each one number from 1 to 100, the probability that they select the same number is:
A. \[\dfrac{99}{100}\]
B. $ \dfrac{1}{100} $
C. $ \dfrac{^{100}{{C}_{2}}}{{{100}^{2}}} $
D. $ \dfrac{100\times 99}{^{100}{{C}_{2}}} $

Answer 1

Hint: Here, we have been given two boys who each select a number from 1 to 100 and we have to find the probability that both the boys choose the same number. For this, we will first find the probability of the first boy selecting any one number out of the given 100 numbers. Then, we will assume the number he selected to be as ‘x’ and then we will find the probability of the second boy choosing the number ‘x’ from the given 100 numbers. Thus, we will get our answer.

Complete step by step answer:
Here, we have been given two boys who each select a number from 1 to 100.
Now, the first boy can select any one number out of the 100 total numbers. We know that the number of ways of selecting ‘r’ objects out of ‘n’ objects is given by the formula $ ^{n}{{C}_{r}} $ . Here, we have:
n=100
r=1
Thus, the number of ways of selecting any one number out of the numbers from 1 to 100 for the first boy is given as:
 $ ^{100}{{C}_{1}} $
Now, solving it we get:
 $ \begin{align}
  & ^{100}{{C}_{1}} \\
 & \Rightarrow \dfrac{100!}{\left( 100-1 \right)!1!} \\
 & \Rightarrow \dfrac{100!}{99!1!} \\
 & \Rightarrow \dfrac{100\times 99!}{99!\times 1} \\
 & \Rightarrow 100 \\
\end{align} $
We also know that the probability of any event is given as:
 $ \text{Probability=}\dfrac{\text{no}\text{. of favourable outcomes}}{\text{total no}\text{. of outcomes}} $
Here, the number of favorable outcomes is equal to the number of ways any 1 number can be selected out of the numbers from 1 to 100, i.e. 100 as we calculated above and the total number of outcomes is equal to the total numbers which can be selected, i.e. 100.
Thus, the probability of selecting any one number from 1 to 100 for the first boy is given as:
 $ \begin{align}
  & \text{Probability=}\dfrac{\text{no}\text{. of favourable outcomes}}{\text{total no}\text{. of outcomes}} \\
 & \Rightarrow \text{Probability=}\dfrac{100}{100} \\
 & \Rightarrow \text{Probability=1} \\
\end{align} $
Hence, the probability of choosing any 1 number from 1 to 100 for the first boy is 1.
Now, let’s assume that the number the first boy selected from 1 to 100 is ‘x’. Thus, we have to find the probability of the fact that the second boy selects ‘x’ too.
Now, as we know the total numbers the second boy can select are 100 but since he has to select x only, the number of favorable outcomes will be equal to 1.
Thus, the probability that the second boy chooses the same number as the first boy is given as:
 $ \begin{align}
  & \text{Probability=}\dfrac{\text{no}\text{. of favourable outcomes}}{\text{total no}\text{. of outcomes}} \\
 & \Rightarrow \text{Probability=}\dfrac{1}{100} \\
\end{align} $
Hence, the probability that both the boys select the same number from 1 to 100 is $ \dfrac{1}{100} $ .
Thus, option (B) is the correct option.

Note:
Here, the probability of the first boy choosing any number is 1. So, the answer is directly equal to the probability of the second boy choosing the same number. If it were not 1, the answer would be the product of both the probabilities as both these events have to happen simultaneously.