Question

In the word "UNIVERSITY", find the probability that the two "I"s do not come together.

• A. \( \frac{7}{11} \)
• B. \( \frac{8}{11} \)
• C. \( \frac{9}{11} \)
• D. \( \frac{10}{11} \)

Hint:

Remember: When letters repeat in a word, adjust the total number of arrangements by dividing by the factorial of the number of repeated letters. To calculate probabilities, consider the favorable and total outcomes.

Answer 1

The Correct Option is B

Step 1: Total number of ways to arrange the letters of the word "UNIVERSITY". The word "UNIVERSITY" consists of 10 letters. The total number of ways to arrange these 10 letters is calculated by considering the repeated letters. The letter "I" repeats twice. Thus, the total number of arrangements is given by: \[ \text{Total arrangements} = \frac{10!}{2!} = \frac{3628800}{2} = 1814400 \] Step 2: Number of ways in which the two "I"s come together. Treat the two "I"s as a single entity or block. Now, we have the following 9 units to arrange: "II", U, N, V, E, R, S, T, Y. Thus, the number of arrangements of these 9 units is: \[ \text{Arrangements with "I"s together} = 9! = 362880 \] Step 3: Number of ways in which the two "I"s do not come together. The number of ways in which the two "I"s do not come together is the total number of arrangements minus the number of arrangements where the "I"s are together: \[ \text{Arrangements with "I"s not together} = 1814400 - 362880 = 1451520 \] Step 4: Probability that the two "I"s do not come together. The probability is the ratio of favorable outcomes (where the two "I"s do not come together) to the total outcomes (total arrangements): \[ \text{Probability} = \frac{\text{Arrangements with "I"s not together}}{\text{Total arrangements}} = \frac{1451520}{1814400} = \frac{8}{11} \] Answer: Therefore, the probability that the two "I"s do not come together is \( \frac{8}{11} \).