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If the letters in the word BOOLEAN are arranged at random, what is the probability that the two O's remain together in the arrangement?

Short Answer

Expert verified
The probability that when the letters in the word BOOLEAN are arranged randomly, the two O's will stay together is \( \frac{{6! \cdot 2!}}{{\frac{{7!}}{{2!}}}} \)

Step by step solution

01

Calculate total permutations

The first step is to determine the total permutations of the word BOOLEAN without any conditions. As there are 7 letters in the word, it has 7! permutations. But since the letter O is repeated twice, we divide by 2! to avoid counting identical arrangements twice. So total permutations \(P_{total}\) are \( P_{total} = \frac{{7!}}{{2!}} \)
02

Calculate permutations keeping two O's together

If we want the two O's to stay together, we can treat them as one letter. Then the word BOOLEAN becomes BO(O)LEAN, with 6 distinct entities. The permutations are then 6!. But don't forget to multiply by 2! as the two O's in the pair can switch positions. So permutations keeping two O's together \(P_{O's\_together}\) are \( P_{O's\_together} = 6! \cdot 2! \)
03

Calculate the probability

The probability is the ratio of the desired outcomes (permutations keeping two O's together) to the total possible outcomes (total permutations). Thus, the probability \(P\) is \( P = \frac{{P_{O's\_together}}}{{P_{total}}} = \frac{{6! \cdot 2!}}{{\frac{{7!}}{{2!}}}} \)

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Key Concepts

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

Permutations
Permutations are a way to organize or arrange a set of items where order is important. In simple terms, permutations count how many different ways we can place items in a sequence. For example, if you have the word "BOOLEAN" and want to determine different ways to rearrange its letters, you would calculate its permutations.
The general formula for permutations of a set of items is the factorial of the number of items, denoted as "n!". In the word "BOOLEAN," since there are 7 letters, you might first think there are 7! ways to permute them.
However, permutations become slightly more complex when identical items are present. This is why understanding permutations is crucial when dealing with repeated elements.
Repeated Elements in Permutations
When calculating permutations for a word with repeated letters, it's essential to account for repetitions to avoid overcounting equivalent arrangements. Here's the key: If you have any letter that repeats, say 'n' times, you divide the total number of permutations by the factorial of the number of times the letter appears. This corrects the calculation by removing duplicates caused by the repeated elements.
In "BOOLEAN," the letter 'O' repeats twice. Thus, instead of 7! permutations, we calculate \[ \frac{7!}{2!} \]
If we didn't adjust for the repeated 'O's, each arrangement where they switch places would seem distinct when actually they are the same.
Understanding how to adjust for repeated elements in permutations is vital in many combinatorial problems.
Combinatorial Probability
Combinatorial probability involves finding the likelihood of a specific arrangement or combination occurring among possible arrangements. It's all about dividing the number of favorable outcomes by the number of total possible outcomes.
In the case of the word "BOOLEAN" and the question of whether the two O's are together, we start by treating the pair of O's as a single unit. This conceptually reduces the size of the problem, allowing us to count fewer permutations, specifically \[ 6! \times 2! \].
The favorable permutations are the ones where the O's stick together.
The final step is to employ probability by dividing these favorable permutations by the total permutations we discussed earlier. This ratio determines the chance of your desired event occurring, providing a clear insight into the probability calculation.
  • Numerator: Permutations with O's together.
  • Denominator: Total permutations, corrected for repetition.
Applying combinatorial probability helps in decoding complex situations with a straightforward arithmetic solution.

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Most popular questions from this chapter

A large jet aircraft has two wheels per landing gear for added safety. The tires are rated so that even with a "hard landing" the probability of any single tire blowing out is only \(0.10\). (a) What is the probability that a landing gear (with two tires) will survive even a hard landing with at least one good tire? (b) In order for the plane to land safely, all three landing gears (the nose and both wing landing gears) must have at least one good tire. What is the probability that the jet will be able to land safely even on a hard landing?

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A professor has two dozen introductory textbooks on computer science and is concerned about their coverage of the topics \((A)\) compilers, \((B)\) data structures, and \((C)\) operating systems. The following data are the numbers of books that contain material on these topics: \(\begin{array}{lll}|A|=8 & |B|=13 \quad|C|=13 \\ |A \cap B|=5 & |A \cap C|=3 & |B \cap C|=6 \\ |A \cap B \cap C|=2 & & \end{array}\) (a) How many of the textbooks include material on exactly one of these topics? (b) How many do not deal with any of the topics? (c) How many have no material on compilers?

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