Question

Letters in the word HULULULU are rearranged. The probability of all three L being together is

• A. $\frac{3}{20}$
• B. $\frac{2}{5}$
• C. $\frac{3}{28}$
• D. $\frac{5}{23}$

Hint:

Probability refers to the likelihood of a specific event occurring out of all possible outcomes.

Answer 1

The Correct Option is C

Given words, HULULULU
\(\therefore\) Total number of sample space \(=\frac{8 !}{4 ! 3 !}\)
and total number of ways all three \(L\) being together \(=\frac{6 !}{41}\)
Required probability \(=\frac{\frac{6 !}{4!}}{\frac{8 !}{4 ! 3 !}}\)
\(=\frac{6! \times 3!}{8 !}=\frac{3}{28}\)

Discover More From Chapter: Permutations and Combinations

Answer 2

The Correct Answer is (C)

Real Life Applications

Some real-life applications of permutations and combinations are 
1. They are used in cryptography to create secure encryption algorithms. 
2. Permutations and combinations are used in genetics to study the inheritance of genes. 
3. It is used in sports to calculate the odds of winning a game or tournament. 
4. Permutations and combinations are used in engineering to design safe and efficient structures.

Question can also be asked as

1. What is the probability that the three L's will be together if the letters in the word HULULULU are rearranged randomly? 
2. What is the probability of flipping a coin three times and getting heads all three times? 
3. What is the probability of randomly choosing a day of the week and getting a weekday? 
4. What is the probability of randomly choosing a number from 1 to 100 and getting a multiple of 5?

 

Answer 3

The Correct Answer is (C)

In probability theory, the concept of arranging letters in a word and calculating the likelihood of specific events occurring is important. One such scenario involves rearranging the letters in the word "HULULULU" and determining the probability of all three L's appearing together. 

Probability and Letter Arrangement

• Probability refers to the likelihood of a specific event occurring out of all possible outcomes.
• When rearranging the letters in a word, different arrangements can result in various patterns and occurrences.

Analyzing the Word "HULULULU"

• The word "HULULULU" contains three L's.
• We are interested in determining the probability of all three L's being adjacent to each other in the rearranged word.

Total Possible Arrangements

• To calculate probabilities, we need to consider the total number of possible arrangements for the letters in the word.
• The word "HULULULU" consists of eight letters, so the total number of arrangements is 8!

Calculating the Favorable Outcomes

• For the three L's to be together, we can treat them as a single unit or block in the rearranged word.
• Now, we have six units: (L), H, U, (L), U, (L), U.
• The three L's can be arranged within this block in 3! (factorial) ways.

Probability Calculation

• To calculate the probability, we divide the number of favourable outcomes (arrangements with the three L's together) by the total possible outcomes.
• The probability of the three L's being together is given by: P = (Number of favourable outcomes) / (Total possible outcomes).

Simplifying the Probability

• The number of favorable outcomes is 3! since the three L's can be arranged among themselves in 3! ways.
• The total possible outcomes are 8! since there are eight letters to rearrange.
• Thus, the probability of all three L's being together is P = 3! / 8!.

In the rearranged word "HULULULU," the probability of all three L's being adjacent to each other is calculated to be 3/28.

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Answer 4

The word "HULULULU" consists of 8 letters where H, U, and L are repeated:
H appears 1 time, U appears 4 times, L appears 3 times
The formula for permutations of a multiset is given by:
\(\frac{n!}{n_1! \times n_2! \times \ldots \times n_k!}\)

For "HULULULU":

\(\frac{8!}{1! \times 4! \times 3!}\)

\(8! = 40320, \quad 1! = 1, \quad 4! = 24, \quad 3! = 6\)
\(\frac{40320}{1 \times 24 \times 6} = \frac{40320}{144} = 280\)

Number of favorable permutations (with all three L's together):
Consider "LLL" as a single entity or block. Thus, we treat "LLL" as one letter. Now we have the following blocks: {LLL}, H, U, U, U, U.
This gives us 6 entities to arrange: {LLL, H, U, U, U, U}.
The number of permutations of these 6 entities, where U repeats 4 times, is given by:

\(\frac{6!}{4!}\)
\(6! = 720, \quad 4! = 24\)

\(\frac{720}{24} = 30\)

The probability is the number of favorable permutations divided by the total number of permutations:
\(\frac{30}{280} = \frac{3}{28}\)

So, the correct option is (C): \(\frac{3}{28}\)