Key Concepts and Formulas

• 1. Probability of an Event: The probability of an event $E$ is defined as the ratio of the number of favorable outcomes $n(E)$ to the total number of possible outcomes $n(S)$ in the sample space. $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{n(E)}{n(S)}$
• 2. Permutations of a Multi-set: The number of distinct arrangements (permutations) of $n$ objects, where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type k, is given by the formula: $\frac{n!}{n_1! n_2! \dots n_k!}$
• 3. Divisibility Rule for 2: A whole number is divisible by 2 if and only if its last digit (units digit) is an even digit (0, 2, 4, 6, or 8).

Step-by-Step Solution

Step 1: Understand the Problem and Identify Given Information We are asked to find the probability that a seven-digit number, formed using specific digits, is divisible by 2. The given digits are: 3, 3, 4, 4, 4, 5, 5. Let's count the frequency of each digit:

• Digit '3': 2 times
• Digit '4': 3 times
• Digit '5': 2 times The total number of digits is $n = 2 + 3 + 2 = 7$.

Step 2: Calculate the Total Number of Possible Outcomes (Sample Space), $n(S)$

• What we are doing: We need to find the total number of distinct seven-digit numbers that can be formed using the given set of digits. This represents our sample space $S$.
• Why we are doing this: This value will be the denominator in our probability calculation. Since some digits are repeated, we must use the formula for permutations of a multi-set to avoid counting identical numbers multiple times.
• Calculation: We have 7 digits in total, with '3' repeated 2 times ($n_3=2$), '4' repeated 3 times ($n_4=3$), and '5' repeated 2 times ($n_5=2$). Applying the multi-set permutation formula: $n(S) = \frac{7!}{n_3! n_4! n_5!} = \frac{7!}{2! \cdot 3! \cdot 2!}$ Let's calculate the factorials: $7! = 5040$ $2! = 2$ $3! = 6$ $2! = 2$ Substitute these values: $n(S) = \frac{5040}{2 \cdot 6 \cdot 2} = \frac{5040}{24}$ $n(S) = 210$ So, there are 210 distinct seven-digit numbers that can be formed.

Step 3: Calculate the Number of Favorable Outcomes, $n(E)$

• What we are doing: We need to find the number of these seven-digit numbers that are divisible by 2. This represents our favorable event $E$.
• Why we are doing this: This value will be the numerator in our probability calculation. We apply the divisibility rule for 2 to identify which numbers satisfy the condition.
• Applying the Divisibility Rule: A number is divisible by 2 if its last digit is even. From our set of given digits {3, 3, 4, 4, 4, 5, 5}, the only even digit available is '4'. Therefore, for the number to be divisible by 2, its last digit must be '4'.
• Forming the Favorable Numbers:
  1. Fix the last digit: Place one '4' in the units place. _ _ _ _ _ _ 4
  2. Determine the remaining digits: After using one '4', we are left with 6 digits to arrange in the first 6 positions. The original digits were: {3, 3, 4, 4, 4, 5, 5} The remaining digits are: {3, 3, 4, 4, 5, 5}
  3. Count frequencies of remaining digits:
     • Digit '3': 2 times ($n'_3=2$)
     • Digit '4': 2 times (since one '4' was used, $n'_4=2$)
     • Digit '5': 2 times ($n'_5=2$) The total number of remaining digits is $n' = 6$.
• Calculation: Now, we calculate the number of ways to arrange these 6 remaining digits using the multi-set permutation formula: $n(E) = \frac{6!}{n'_3! n'_4! n'_5!} = \frac{6!}{2! \cdot 2! \cdot 2!}$ Let's calculate the factorials: $6! = 720$ $2! = 2$ Substitute these values: $n(E) = \frac{720}{2 \cdot 2 \cdot 2} = \frac{720}{8}$ $n(E) = 90$ So, there are 90 distinct seven-digit numbers (formed from the given digits) that are divisible by 2.

Step 4: Calculate the Probability, $P(E)$

• What we are doing: We combine the total number of outcomes and the number of favorable outcomes to find the probability.
• Why we are doing this: This is the final step in applying the fundamental definition of probability.
• Calculation: Using the probability formula: $P(E) = \frac{n(E)}{n(S)}$ Substitute the values calculated in Step 2 and Step 3: $P(E) = \frac{90}{210}$ Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 30: $P(E) = \frac{90 \div 30}{210 \div 30} = \frac{3}{7}$ Alternatively, and more efficiently for competitive exams, we can simplify using factorial expressions directly: $P(E) = \frac{\frac{6!}{2!2!2!}}{\frac{7!}{2!3!2!}}$ Rewrite as a multiplication and cancel common terms ($2!$): $P(E) = \frac{6!}{2!2!2!} \times \frac{2!3!2!}{7!} = \frac{6! \times 3!}{7! \times 2!}$ Expand $7! = 7 \times 6!$ and $3! = 3 \times 2 \times 1 = 6$, and $2! = 2 \times 1 = 2$: $P(E) = \frac{6! \times 6}{(7 \times 6!) \times 2}$ Cancel $6!$ from the numerator and denominator: $P(E) = \frac{6}{7 \times 2} = \frac{6}{14}$ Simplify by dividing numerator and denominator by 2: $P(E) = \frac{3}{7}$

Common Mistakes & Tips

• Ignoring Repeated Digits: A common error is to calculate permutations as $n!$ without accounting for repeated digits. Always use the multi-set permutation formula when objects are not all distinct.
• Misapplication of Divisibility Rules: Ensure you correctly identify the condition for the favorable event. For divisibility by 2, only the units digit matters.
• Calculation Errors with Factorials: Factorials can become large quickly. To minimize errors, especially in fractions involving factorials, try to simplify by canceling common terms before performing multiplications. For example, $\frac{n!}{(n+1)!} = \frac{1}{n+1}$.
• Fixed Positions: When a condition fixes certain positions (like the last digit for divisibility), ensure you reduce the set of available digits and the total number of positions for the remaining arrangement calculation.

Summary

To find the probability that a seven-digit number formed from the given digits {3, 3, 4, 4, 4, 5, 5} is divisible by 2, we first calculated the total number of distinct seven-digit numbers possible using the multi-set permutation formula, yielding $n(S) = 210$. Next, we determined the number of favorable outcomes by fixing the last digit as '4' (the only even digit available) and then arranging the remaining six digits, which gave $n(E) = 90$. Finally, we divided the number of favorable outcomes by the total number of outcomes, $P(E) = \frac{90}{210} = \frac{3}{7}$.

The final answer is $\boxed{\frac{3}{7}}$, which corresponds to option (D).