Answer Show  Hint: For finding the probability of getting a consonant, we can use the equation of probability. Since probability is the number of favourable in total number, knowing the number of consonants and total number of letters in the alphabet, we can find the answer. Formula used: Given that a letter of English alphabet is chosen at random. We have to find the probability that the chosen letter is a consonant. Consonants are letters other than vowels in the alphabet. In English alphabet we have five vowels which are “a, e, i, o & u”. Since there are $26$ letters in total, we get the number of consonants as $26 - 5 = 21$. Probability of an event is obtained by dividing the number of favourable outcomes by total number of outcomes. So here the probability of getting a consonant is found by dividing the number of consonants by the total number of letters in English alphabet. If $C$ is the event of getting consonant we have, $P(C) = \dfrac{{21}}{{26}}$ $\therefore $ The probability that the chosen letter is consonant is $\dfrac{{21}}{{26}}$. Note: We can also solve the problem in another way. We know the sum of probabilities is equal to one. When choosing a letter from English alphabet at random, there are only two possibilities; either vowel or consonant. Since there are five vowels, the probability of getting a vowel is $\dfrac{5}{{26}}$. So the probability of getting consonant is $1 - \dfrac{5}{{26}} = \dfrac{{21}}{{26}}$.
Register now for special offers +91 Home > English > Class 12 > Maths > Chapter > Permutations And Combinations > How many different words with ... UPLOAD PHOTO AND GET THE ANSWER NOW! Text Solution Solution : The word 'Jungle Contains 2 vowels and 4 consonants. Each word contains one vowel and one consonant.<br> The number of different words formed.<br>`2xx""^4P_1xx""^2P_1=2xx4xx2=16` Answer Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Add a public comment... Follow Us: Popular Chapters by Class: Before we can start calculating the probability that two letters picked at random from a ciphertext are identical we should learn how to calculate probabilities in general! The probability of an event occurring is \(\frac{r}{n}\) where \(n\) is the total number of possible
outcomes and the event can happen in \(r\) equally likely ways. Given the five letters, A, B, C, D, E. If we pick one letter at random what is the probability that it is an A? a vowel? a consonant? Since there is only one A, the probability of picking an A is 1/5. Since there are two vowels, the
probability of picking a vowel is 2/5. Since there are three consonants, the probability of picking a consonant is 3/5. Note there is another way to calculate this. Since the only way a letter can be a consonant is if it is not a vowel, we could calculuate this as the probability of not getting a vowel. The probability of not getting a vowel is 1-2/5=3/5. If the probability of an event is \(P(A)\text{,}\) then the probability of event \(A\) not occurring is \(1-P(A)\text{.}\) That is, \(P(\text{not } A)=1-P(A)\text{.}\) Suppose we pick two letters at random from the set A, B, C, D, E. What is the probability that both of them are vowels? There are
different ways to think about this depending on our assumptions. Are we picking one letter and replacing it so we may pick the same letter both times, or are we picking both letters at the same time? If we are picking both letters at the same time do we care about the order in which we got the letters or do we not care about the order? Let's look at all three of those cases. Case 1: Replacement. We pick one letter and return it so that picking exactly the
same letter twice is possible. In this case we may pick AA, AE, EA, or EE to pick two vowels. There are 25 ways to pick any two letters from this set, namely AA, AB, AC, AD, AE, BA, BB, BC, BD, BE, CA, CB, CC, CD, CE, DA, DB, DC, DD, DE, EA, EB, EC, ED, EE. Thus, the probability is 4/25. Case 2: No replacement and order does matter. In this case, we pick both letters at the same time. Thus, we may not pick the exact same letter twice. The possible ways to
pick two vowels are AE or EA. There are 20 ways to pick any two letters from this set, namely, AB, AC, AD, AE, BA, BC, BD, BE, CA, CB, CD, CE, DA, DB, DC, DE, EA, EB, EC, ED. Thus, the probability is 2/20. Case 3: No replacement and order does not matter. In this case, there is only one way to pick two vowels namely A and E. There are now only 10 ways to pick any two letters from this set, namely, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. Thus, the probability
is 1/10. In this case, we get the same probability as in Case 2, but that will not usually happen. Let's talk about how to make the calculation for the denominator in each of these three cases. For Case 1, we have a set of five letters and want to pick two of them allowing replacement. In this case, there are 5 choices for the first letter and five choices for the second letter, so the total number of ways to choose two letters from a set of
5 allowing replacement is \begin{equation*} 5 \times 5 = 25. \end{equation*} This same calculation holds in general. The number of ways to pick \(r\) objects from a set of \(n\) when replacement is allowed is \(n \times n \times \cdots \times n = n^r\text{.}\) Order always matters for replacement. For Case 2 we have a set of five letters and want to pick two of them; we care about order but do not allow replacement. In this case, there are five choices of letters to pick for the first letter, but only four choices of letter to pick for the second letter, since we cannot pick whatever letter we picked first. Thus the total number of ways to choose two letters from a set of 5 where order matters but replacement is not allowed is \begin{equation*} 5 \times 4 = 20. \end{equation*} This type of computation is referred to as a permutation. A permutation \(P(n,r)=\) is the number of ways to arrange \(r\) objects from a set of \(n\text{.}\) (Order matters.) Note that order matters for a permutation because we are arranging the objects. In general, to compute \(P(n,r)\) there are \(n\) ways to pick the first object, then \(n-1\) ways to pick the second object and so on down to \(n-r+1\) ways to pick the \(r\)th object. We can rewrite this using factorials to get: \(P(n,r)=n\times(n-1)\times \cdots \times (n-r+1) = \frac{n!}{(n-r)!}\) For Case 3 we have a set of five letters and want to pick two of them; we do not care about order and do not allow replacement. In this case our calculation of \(5 \times 4 = 20 \) is much too big and overcounts many pairs of letters. For example, the the letters A and B, it counts both AB and BA separately. In fact, it double counts for every pair of letters in exactly the same way. Thus the total number of ways to choose two letters from a set of 5 where order does not matter and replacement is not allowed is \begin{equation*} \frac{5 \times 4}{2} = 10. \end{equation*} This is referred to as a combination. A combination \(C(n,r)=\) is the number of ways to choose \(r\) objects from a set of \(n\) objects. (Order does not matter.) Note that order does not matter for a combination because we are choosing objects but not arranging them. In this case, the permutation has overcounted by all the possible orderings of \(r\) objects. Thus we need to divide the permutation by \(r!\text{.}\) \(C(n,r)=\frac{P(n,r)}{r!} = \frac{n!}{(n-r)!r!}\) Time for you to explore permutations, combinations and probability in Investigation: Probability Practice. Let's look at an example where the ideas are a little more complicated. We have a bucket of 20 letters. These letters include A, B, C and are either red or blue. There are 4 red As, 6 blue As, 4 red Bs, 3 blue Bs, 1 red Cs and 2 blue Cs. In this case, we do not care about order when we are picking letters.
Solution
Both Investigation: Probability Practice and Example 3.2.10 should have raised some important questions.
For the first question, if the two outcomes are disjoint (meaning they can't both happen at the same time) then we can add the probabilities. Often, this corresponds to a probability involving “or”. For example, when we are picking two letters from the bucket, picking AA or BB, are disjoint events. We can't pick AA and BB at the same time. If the number of ways event A can happen is \(r\) and the number of ways event B can happen is \(s\) and event A and event B are disjoint events, then the number of ways event A or event B can happen is \(r+s\text{.}\) For the second question, if the two outcomes are independent from each other and we need both of them to occur, then we multiply probabilities. For example, when picking one letter as an A and the other letter as a B we multiplied the probabilities because picking an A for one letter does not impact the ways we can choose B as a second letter. These events are independent. Often, this corresponds to a probability involving “and”. However, this is not always the case. The events may not be independent such as in the example of picking one A and the other letter being red. It is also possible that “and” is used in a different way. For the probability of picking one letter that is both an A and red, we would not multiply the number of As times the number of red letters. If the number of ways event A can happen is \(r\) and for each way event A can happen, the number of ways event B can happen is \(s\text{,}\) then the number of ways event A and event B can happen is \(r \times s\text{.}\) The last question doesn't have an easy answer. You'll just have to think about it for each problem. Sometimes it is easier to compute the opposite event, \(p\text{,}\) and use \(1-p\) for the probability of the desired event. This also points out that probabilities should be between 0 and 1. If a probability is 0 then that event will never happen and if a probability is 1 then that event is guaranteed to happen. How many ways the consonant and vowel?Note- In English alphabet set we have in total 27 alphabets which can broadly be classified into two categories that are vowels and consonants. (A, E, I, O, U) are the vowels while rest are marked as consonants.
How many possible outcomes are there in the English consonants?Complete step-by-step answer:
Consonants are letters other than vowels in the alphabet. In English alphabet we have five vowels which are “a, e, i, o & u”. Since there are 26 letters in total, we get the number of consonants as 26−5=21.
How many possible outcomes are there in the English vowels?There are three sections with vowels: those with the letters A, E, and I. We already said there are eight sections in total. Hence, there are eight possible outcomes.
How many outcomes are there in getting a vowel letter in the English alphabet?Answer: 5/26. Assuming that only 26 alphabets (a to z) are considered & there are 5 vowels a,e,I,o,u.
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