Question

If one letter is chosen at random from the word ‘mathematics’. Find the probability of choosing:
(A). ‘m’ alphabet
(B). ‘a’ alphabet
(C). ‘c’ alphabet

Answer 1

Hint: Use the mathematical definition of probability, which is given as:
$P=\dfrac{\text{Number of favourable cases}}{\text{Total cases}}$
Count the total number of letters/alphabets in the word “mathematics” and put it as total cases in the above relation. Now, get the total number of alphabet ‘m’ for number of favorable cases to the relation and get the answer of the first part. Similarly, get the answer for the other two parts by putting favorable cases as the total number of alphabets ‘a’ for the second part and ‘c’ for the third part.

Complete step-by-step solution -
As we know probability can be defined mathematically as:
$P=\dfrac{\text{Number of favourable cases}}{\text{Total cases}}$………………………… (1)
Now, coming to the question, it is given that we have to choose one letter at random from the word “mathematics” and hence, we need to determine the probability of choosing
‘m’ alphabet
‘a’ alphabet
‘c’ alphabet
Now, we are given the word “mathematics”. So, let us write down the different number of alphabets with their counting in the word. So, we get:
$\begin{align}
  & m\to 2 \\
 & a\to 2 \\
 & t\to 2 \\
 & h\to 1 \\
 & e\to 1 \\
 & i\to 1 \\
 & c\to 1 \\
 & s\to 1 \\
\end{align}$
So, the total number of alphabets in the given word is given as $=11$.
Now, coming to the question:-
(A). ‘m’ alphabet
Here, we need to find the probability for choosing the ‘m’ alphabet from the given word “mathematics”. So, we can get a probability using relation (1), where we know
Total number of cases = Total number of alphabets = 11.
Favorable cases = number of ‘m’ alphabet in given word = 2
So, we get the probability as:
$P=\dfrac{2}{11}$
(B). ‘a’ alphabet
Here, we need to find the probability for choosing the alphabet ‘a’ from the given word “mathematics”. So, we can get the probability using relation (1), where we know:
Total number of cases = Total number of alphabets = 11.
Favorable cases = number of ‘a’ alphabet in given word = 2
So, we get the probability to choose the alphabet ‘a’ as:
$P=\dfrac{2}{11}$
(C). ‘c’ alphabet
Here, we need to find the probability for choosing the alphabet ‘c’ from the given word “mathematics”. So, we can get probability using relation (1), where we know:
Total number of cases = Total number of alphabets = 11.
Favorable cases = number of ‘c’ alphabet in given word = 1
So, we get the probability to choose the alphabet ‘c’ as:
$P=\dfrac{1}{11}$.

Note: Be careful with the counting of letters, don’t miss any of the letters required for the relation to get the required probability. So, take care of these kinds of problems.
The probability of any event cannot be more than 1. So, if you are getting the probability of given events more than 1, then you can apply the wrong formula. So, be careful with this concept as well, it will give you verification of the solution.