Question

If 4 letters are selected at random from the letters of the word PROBABILITY, compute the probability that at least one letter is repeated in the selection.

• A. \(\frac{43}{170}\)
• B. \(\frac{19}{61}\)
• C. \(\frac{57}{184}\)
• D. \(\frac{29}{155}\)

Hint:

When calculating the probability of at least one occurrence of an event, consider using the complement rule, which simplifies the process by subtracting the probability of the event not happening from 1.

Answer 1

The Correct Option is B

Step 1: Analyze the composition of the word "PROBABILITY". The word "PROBABILITY" consists of 11 letters: P, R, O, B, A, B, I, L, I, T, Y. We observe that the letters B and I are repeated. Thus, the distinct letters are: P, R, O, A, B, I, L, T, Y, making 9 distinct letters in total. 

 Step 2: Calculate the total number of ways to choose 4 letters from 11. The total number of ways to select 4 letters out of the 11 letters (considering repetitions) is: \[ \text{Total ways} = \binom{11}{4} \] Using the formula for combinations: \[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \] 

Step 3: Calculate the number of ways to select 4 letters with no repetition. To select 4 letters with no repetition, we choose from the 9 distinct letters (P, R, O, A, B, I, L, T, Y): \[ \text{Ways with no repetition} = \binom{9}{4} \] Using the formula for combinations: \[ \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

 Step 4: Apply the complement rule to calculate the probability of at least one repetition. The probability of no repetition is: \[ \text{Probability of no repetition} = \frac{\binom{9}{4}}{\binom{11}{4}} = \frac{126}{330} \] Now, the probability of at least one repetition is the complement of the probability of no repetition: \[ \text{Probability of at least one repetition} = 1 - \frac{\binom{9}{4}}{\binom{11}{4}} = 1 - \frac{126}{330} \] \[ = 1 - \frac{63}{165} = \frac{165 - 63}{165} = \frac{102}{165} \] Simplifying the fraction: \[ \frac{102}{165} = \frac{19}{61} \] Thus, the probability that at least one letter is repeated is \( \boxed{\frac{19}{61}} \).