What is the probability of drawing a consonant from the letter of the word?

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.

  1. 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.

  2. 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.

  3. 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.

Example 3.2.10.

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.

  1. Suppose we pick one letter from the bucket.

    1. What is the probability of picking one A or one B?

    2. What is the probability of picking one A or one red letter?

  2. Suppose we pick two letters from this bucket at the same time.

    1. What is the probability of picking one A and one B?

    2. What is the probability that both are the same letter? That is, what is the probability that we pick AA or BB or CC?

    3. What is the probability that at least one letter is blue?

    4. What is the probability that one letter is A and the other letter is red.

Solution

  1. There are \(C(20,1)=20\) ways to pick one letter from the bucket.

    1. There are \(C(10,1)=10\) ways to pick an A and there are \(C(7,1)=7\) ways to pick a B. Since picking an A or picking a B are disjoint events, the probability is \(\frac{C(10,1)+C(7,1)}{20}=\frac{17}{20}\text{.}\)

    2. In this case, there are 10 letter As and 9 red letters, but we cannot add probabilities because there is overlap between A letters and red letters. We need to create multiple cases to eliminate any overlap and avoid double counting the four red As. We could count the number of ways to pick a blue A or pick a red letter instead. This covers all the same cases without double counting. The probability is \(\frac{C(6,1)+C(9,1)}{20}=\frac{15}{20}\text{.}\)

  2. There are \(C(20,2)=190\) ways to pick two letters from the bucket.

    1. Since there is no overlap in picking an A or picking a B, the probability is \(\frac{C(10,1)C(7,1)}{190}=\frac{70}{190}\text{.}\)

    2. Again, there is no overlap in picking two As, two Bs, or two Cs.The probability is \(\frac{C(10,2)+C(7,2)+C(3,2)}{190}=\frac{69}{190}\text{.}\)

    3. There are two ways that at least one letter is blue: either exactly one letter is blue or both letters are blue. We could compute the probably of each of these and add them since they are disjoint cases, but its easier to compute the opposite event. That is, what is the probability that we don't get any blue letters? This means both letters must be red so the probability is\(\frac{C(9,2)}{C(20,2)}\text{.}\) Since this is the event we don't want, the probability of at least one blue is \(1-\frac{C(9,2)}{C(20,2)}=\frac{154}{190}\text{.}\)

    4. There is overlap in the letters that are As and the red letters, so we have to separate into cases to avoid the overlap. We first separate based on whether the A is red or blue. However, if the A is red then we also have separate based on whether the other letter is a red A, so that there are three cases for this example.

      • Case 1: Both letters are red As. There are \(C(4,2)=6\) ways to pick two red As.

      • Case 2: One letter is a red A and the second letter is a red B or C. There are \(C(4,1)*C(5,1)=20\) ways to pick this.

      • Case 3: One letter is a blue A and the second letter is red (any of A, B, C). There are \(C(6,1)*C(9,1)=54\) ways to pick this

        What is the probability of drawing a consonant letter?

        There is a . 19 or 19% chance of choosing a vowel whereas there is a . 81 or 81% chance of choosing a consonant. We have to be careful not to double count probabilities.

        What is the probability of getting a consonant from the word probability?

        $(1)$ Since, we know that the total letters in the word PROBABILITY are 11. Therefore, the probability of choosing a vowel (P)$ = \dfrac{4}{{11}}$. $(2)$ Since, we know that the total letters in the word PROBABILITY are 11. Therefore, the probability of choosing a consonant (P')$ = \dfrac{7}{{11}}$.

        What is the probability of drawing a letter P?

        For a random draw, the probability of drawing an S is 4/11, and the probability of drawing a P is 2/11.