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Chapter 4: Problem 28

Two letters are taken at random from the word HOME. Find the probability that both the letters are vowels. (a) \(1 / 6\) (b) \(1 / 12\) (c) \(3 / 8\) (d) None of these

Short Answer

Expert verified
The probability is \(1/6\), which is option (a).

Step by step solution

01

Identify Total Possible Outcomes

The word 'HOME' consists of 4 letters: H, O, M, E. We are to choose 2 letters from these 4. The number of ways to choose 2 letters out of 4 is calculated using the combination formula: \( \binom{4}{2} \). Calculate this to find the total number of possible outcomes.
02

Calculate Total Possible Combinations

Using the combination formula \( \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \). Therefore, there are 6 possible combinations when choosing 2 letters from the word 'HOME'.
03

Identify Favorable Outcomes

The vowels in 'HOME' are O and E. To get both letters as vowels, we only have the combination OE. Since OE is the only pair of vowels, there is only 1 favorable outcome.
04

Calculate Probability of Favorable Outcomes

The probability of both letters being vowels is the number of favorable outcomes divided by the total number of possible outcomes. So, the probability is \( \frac{1}{6} \).
05

Determine the Correct Option

The probability we calculated is \( \frac{1}{6} \), corresponding to option (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinations
When determining how many ways we can select items from a set, we often use the concept of combinations. Unlike permutations, where the order matters, combinations care only about the selection itself. In our exercise, we want to choose 2 letters from the word "HOME," which has 4 letters in total.

To find the number of combinations, we use the combination formula, which is written as:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
where:
  • \(n\) is the total number of items to choose from
  • \(k\) is the number of items to choose
For "HOME," we select 2 letters from 4, so \( \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \). There are 6 different combinations of letters available when selecting 2 from the whole word.
Vowels
Vowels are important elements in language and have specific roles in linguistics. In the context of our exercise, identifying vowels is crucial because they are the focus for finding the probability. The letters in the word "HOME" are: H, O, M, and E.

The vowels among these are O and E. Understanding which letters are vowels allows us to pinpoint which combinations are favorable when calculating probabilities. Only the pair OE represents a combination of vowels from "HOME." So, when asked how many ways we can pick two letters where both are vowels from "HOME," the answer is just one way.
Discrete Mathematics
Discrete mathematics provides the foundation for understanding concepts such as probability, which in turn helps us solve problems related to combinations. In this particular problem, we apply discrete mathematics to determine the probability of picking two vowels.

The probability of an event is calculated using:
  • Probability = \( \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
Based on the combinations identified in earlier sections:
  • Total possible outcomes = 6
  • Favorable outcomes (all vowels) = 1
Using these, we calculate:
  • Probability = \( \frac{1}{6} \)
This process uses discrete math to break down problems into countable steps and ensures clear solutions to probability questions.

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Most popular questions from this chapter

Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) 'Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) '

Two cards are drawn successively with replacement from well-shuffled pack of 52 cards. Find the probability distribution of the number of aces. [CBSE-1995, 2001]

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is (a) \(8 / 9\) (b) \(7 / 9\) (c) \(2 / 9\) (d) \(1 / 9\)

There are 3 red and 5 black balls in bag. \(A\) and 2 red and 3 black balls in bag \(B\). One ball is drawn from bag \(A\) and two from bag \(B\). Find the probability that out of the 3 balls drawn, 1 is red and 2 are black. [CBSE-96 (C)]

Out of \((2 n+1)\) consecutively numbered tickets, three are drawn at random. The chance that the numbers on them are in A.P. is (a) \(4 n^{2}-1 / 3 n\) (b) \(4 n^{2}+1 / 3 n\) (c) \(3 n / 4 n^{2}-1\) (d) \(3 n / 4 n^{2}+1\) (b) If the smallest number is 1 , the groups of three numbers in A.P. are as \(1,2,3 ; 1,3\), \(5 ; 1,4,7 ; \ldots ; 1, n+1,2 n+1 ;\) and they are \(n\) in number. If the smallest number selected is 2, the possible groupings are \(2,3,4 ; 2,4,6 ; 2,5,8 ; \ldots ; 2, n+1,2 n\) and their number is \(n-1\). If the lowest number is 3, the groupings are \(3,4,5 ; 3,5,7 ; 3,6,9 ; \ldots ; 3, n+2,2 n+1\) their number being \(n-1\). Similarly, it can be seen that if the lowest numbers selected are \(4,5,6,2 n-2,2 n-1\), the numbers of selections, respectively, are \(n-2, n-2, n-3, n-3, \ldots, 2,2,1,1\). Thus, the favourable ways for 2,3 are the same and similarly they are the same for 4,5 and so on. Hence, number of favourably ways $$ \begin{aligned} M &=2(1+2+3+\ldots n-1)+n \\ &=2 \times \frac{(n-1) n}{2}+n=n^{2}-n+n=n^{2} \end{aligned} $$ Also the total number of ways $$ \begin{aligned} N &={ }^{2 n+1} C_{3}=\frac{(2 n+1) \times 2 n \times(2 n-1)}{1 \times 2 \times 3} \\ &=\frac{n\left(4 n^{2}-1\right)}{3} \end{aligned} $$Hence, the required probability $$ =\frac{M}{N}=\frac{3 n^{2}}{n\left(4 n^{2}-1\right)}=\frac{3 n}{4 n^{2}-1} $$
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