Question

From 101 to 500, if a number is chosen at random, then the probability the number end with ‘0’ is
A) \[\dfrac{{41}}{{399}}\]
B) \[\dfrac{{40}}{{399}}\]
C) \[\dfrac{1}{{10}}\]
D) \[\dfrac{1}{2}\]

Answer 1

Hint:
Here, we will first find the total number of possible outcomes from 101 to 500. Then, we will find the sequence of the numbers which end with ‘0’ and are from 101 to 500. We will observe that the sequence forms an Arithmetic Progression whose last term is known by us. Thus, using the formula of last term of an AP, we will be able to find the favourable outcomes. We will then find the required probability of the chosen number being a number which ends with ‘0’.

Formula Used:
Probability, \[P = \]Number of favourable outcomes\[ \div \] Total number of outcomes

Complete step by step solution:
First, we will find the total numbers from 101 to 500
Thus, from 101 to 500, we have
\[500 - 101 = 399\]numbers
Thus, our total outcomes is 399.
Now, we are required to choose a number out of these numbers such that the number end with ‘0’
We know that the numbers which end with ‘0’ and lie from 101 to 500, are:
\[110,120,130,...500\]
If we observe this sequence, then, each term is greater than the preceding term by 10.
Thus, this forms an Arithmetic Progression or an AP.
Here, the first term, \[a = 110\]
Common difference, \[d = \left( {120 - 110} \right) = 10\]
The last term, \[{a_n} = 500\]
Now, substituting the known values in the formula \[{a_n} = a + \left( {n - 1} \right)d\], we get
\[500 = 110 + \left( {n - 1} \right)\left( {10} \right)\]
Subtracting 110 from both sides, we get
\[ \Rightarrow 390 = \left( {n - 1} \right)\left( {10} \right)\]
Dividing both sides by 10, we get
\[ \Rightarrow 39 = \left( {n - 1} \right)\]
Adding 1 on both sides, we get
\[ \Rightarrow n = 40\]
Therefore, there are 40 numbers which end with ‘0’.
Hence, the number of favourable outcomes \[ = 40\]
We know that, Probability, \[P = \] Number of favourable outcomes \[ \div \] Total number of outcomes.
Hence, the probability that the number end with ‘0’ ,\[P = \dfrac{{40}}{{399}}\]
Therefore, then the probability that the number end with ‘0’ is \[\dfrac{{40}}{{399}}\]

Hence, option B is the correct answer.

Note:
Probability is a branch of mathematics which tells us about how likely an event is to occur or the possibility of that particular event being true. The probability of any event always lies between 0 and 1. Where, if the probability of any event is 0 then, it is considered as an impossible event, as for example, if there are only blue balls in a bag, then, the probability of the ball taken out being red is 0 as this is impossible to happen. Similarly, if the probability of any event is 1, then, that particular event is a definite event as in this case only, the probability that the ball was taken out is blue will always remain 1 as this is definite to happen.