Question

A word consists of 9 letters, 5 consonants and 4 vowels. Three letters are chosen at random. Find the probability such that more than one vowel will be selected.

Answer 1

Hint: In order to solve this question, we will first find the number of possible cases of each given conditions by the formula of combination, that is, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. And then, we can solve this question with the formula of probability, that is, $\text{probability=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$.

Complete step-by-step answer:
In this question, we have been given a word of 9 letters in which 5 are consonants and 4 are vowels. And from them, three letters are chosen at random. And we have been asked to find the probability that more than one vowel will be selected.
To solve this question, we should know that to choose r out of n items, formula of combination is applied, that is, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Now, we know that probability of any event is given by the formula, $\text{probability=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$
Now, we have been given the event as choosing 3 letters at random such that more than one vowels will be selected from a word of 9 letters. Now, we know that for total outcomes of the event, we have to choose 3 letters out of 9, irrespective of vowels and consonants. So, we can say,
Total outcomes = $^{9}{{C}_{3}}$
Now, we know that for favourable outcomes, we want 3 letters to be selected such that it contains more than one vowel, that means, out of 3 letters either 2 or 3 vowels must be selected out of 4 present in the word. If 2 vowels will be selected, then 1 consonant will also be selected out of 5. So, we can say,
Favourable outcomes = $^{4}{{C}_{2}}{{\times }^{5}}{{C}_{1}}{{+}^{4}}{{C}_{3}}$
Now, according to the formula of probability, we can say,
Probability of choosing 3 letters such that more than one vowel will be selected, $\text{P=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$
Now, we will put the value of favourable outcomes and the total outcomes. So, we will get,
$P=\dfrac{^{4}{{C}_{2}}{{\times }^{5}}{{C}_{1}}{{+}^{4}}{{C}_{3}}}{^{9}{{C}_{3}}}$
Now, we will simplify it further, so we can write,
\[\begin{align}
  & P=\dfrac{\dfrac{4!}{2!\left( 4-2 \right)!}\times \dfrac{5!}{1!\left( 5-1 \right)!}+\dfrac{4!}{3!\left( 4-3 \right)!}}{\dfrac{9!}{3!\left( 9-3 \right)!}} \\
 & P=\dfrac{\dfrac{4\times 3\times 2!}{2!2!}\times \dfrac{5\times 4!}{1!4!}+\dfrac{4\times 3!}{3!1!}}{\dfrac{9\times 8\times 7\times 6!}{3!6!}} \\
 & P=\dfrac{\dfrac{4\times 3}{2}\times \dfrac{5}{1}+\dfrac{4}{1}}{\dfrac{9\times 8\times 7}{3\times 2}} \\
 & P=\dfrac{6\times 5+4}{3\times 4\times 7} \\
 & P=\dfrac{30+4}{84} \\
 & P=\dfrac{34}{84}=\dfrac{17}{42} \\
\end{align}\]
Hence, the probability of choosing 3 letters out of 9 letters such that more than one vowels will be selected is $\dfrac{17}{42}$.

Note: While solving this question, there are high chances of calculation mistakes, as it had a lot of calculations. Also, we might write the formula of combination wrong. Also, in case of choosing 2 vowels and 1 consonant, we might not include 1 consonant but that will mean that, we want just 2 letters out of 9. So, we have to include 1 consonant in that situation.