Question

Each of the letters of the word 'AUTHORIZES' is written on identical circular disks and put in a bag. They are well shuffled. If a disc is drawn at random from the bag,
what is the probability that the letter is:
i) a vowel
ii) one of the first 9 letters of the English alphabet which appears in the given word
iii) one of the last 9 letters of the English alphabet which appears in the given word?​

Answer 1

Hint : To get answers to this type of question first count the total number of letters in the word AUTHORIZES that number will be the total number sample spaces. Then to answer the first part of the solution we will count the total number of vowels that are present in the word AUTHORIZES. Then use the formula of probability. For the second part we will count how many letters of the first 9 letters of English alphabet. Then apply the formula of probability and get the required answer. Similarly for the third case we will count the number of letters that come till 9 of English alphabet then by applying the formula of probability we will calculate the probability.

Complete step-by-step answer :
Part i) probability of getting a vowel
The total number of different letters in the word AUTHORIZES is 10 in which
Number of vowels are A,U,I,O,E means 5 vowels are present in this word now the total number of consonant are 5 (T,H,R,Z,S)
As we have to find the probability of getting a vowel
We know that, $Probability = \dfrac{\text{number of favourable cases}}{\text{ total number of cases}}$
$\Rightarrow$ So probability of getting a vowel $= \dfrac{5}{{10}}$ or $\dfrac{1}{2}$ .
Part ii) one of the first 9 letters of the English alphabet which appears in the given word are A, E, H, I
Means 4 letters
$\Rightarrow$ So the probability $= \dfrac{4}{{10}}$ or $\dfrac{2}{5}$
Similarly
For part iii) one of the last 9 letters of the English alphabet which appears in the given word = 5 letters in this word
$\Rightarrow$ So the probability $= \dfrac{5}{{10}}$ or $\dfrac{1}{2}$
Hence the probabilities in each parts are respectively, $\dfrac{1}{2}$ , $\dfrac{2}{5}$ and $\dfrac{1}{2}$

Note : In this problem concept is quite simple only need to pay attention to counting the letters among the first 9 letters of English alphabet and the last 9 letters of English alphabet. If we are mistaken in counting we will get the wrong answer.