A 4-digit number is selected at random formed by using the digits 0, 1, 2, 3 and 4 (where repetition of digits is not allowed).
1/4
Understanding the Problem
We need to find the probability that a 4-digit number, formed using the digits 0, 1, 2, 3, 4 without repetition, is divisible by 6.
A number is divisible by 6 if it's divisible by both 2 and 3.
- Divisibility by 2: The last digit must be even (0, 2, or 4).
- Divisibility by 3: The sum of the digits must be divisible by 3.
Calculating Total Possible 4-Digit Numbers
The digits available are {0, 1, 2, 3, 4}. A 4-digit number cannot start with 0.
- The first digit has 4 choices (1, 2, 3, 4).
- The remaining 3 digits can be chosen and arranged from the remaining 4 digits in \(P(4, 3)\) ways.
- \(P(4, 3) = \frac{4!}{(4-3)!} = 4 \times 3 \times 2 = 24\).
- Total possible 4-digit numbers = \(4 \times 24 = 96\).
Calculating Favorable Outcomes (Divisible by 6)
First, find sets of 4 digits from {0, 1, 2, 3, 4} whose sum is divisible by 3.
- Set 1: {0, 1, 2, 3}. Sum = 6 (divisible by 3).
- Set 2: {0, 2, 3, 4}. Sum = 9 (divisible by 3).
Now, count the 4-digit numbers formed using these sets that are also divisible by 2.
Case 1: Digits {0, 1, 2, 3}
The number must end in 0 or 2.
- Ending in 0: The first three digits are permutations of {1, 2, 3}. Number of ways = 3! = 6.
- Ending in 2: The last digit is 2. The remaining digits are {0, 1, 3}. The first digit cannot be 0.
- Total permutations of {0, 1, 3} for the first three places = 3! = 6.
- Permutations starting with 0 (e.g., 0132) are not 4-digit numbers. There are 2! = 2 such numbers (0132, 0312).
- Valid numbers ending in 2 = 6 - 2 = 4.
- Total numbers divisible by 6 from this set = 6 + 4 = 10.
Case 2: Digits {0, 2, 3, 4}
The number must end in 0, 2, or 4.
- Ending in 0: The first three digits are permutations of {2, 3, 4}. Number of ways = 3! = 6.
- Ending in 2: The last digit is 2. The remaining digits are {0, 3, 4}. The first digit cannot be 0.
- Total permutations of {0, 3, 4} = 3! = 6.
- Permutations starting with 0 (e.g., 0342) = 2! = 2.
- Valid numbers ending in 2 = 6 - 2 = 4.
- Ending in 4: The last digit is 4. The remaining digits are {0, 2, 3}. The first digit cannot be 0.
- Total permutations of {0, 2, 3} = 3! = 6.
- Permutations starting with 0 (e.g., 0234) = 2! = 2.
- Valid numbers ending in 4 = 6 - 2 = 4.
- Total numbers divisible by 6 from this set = 6 + 4 + 4 = 14.
Total Favorable Outcomes
Total numbers divisible by 6 = Numbers from Case 1 + Numbers from Case 2 = 10 + 14 = 24.
Calculating the Probability
Probability = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)
Probability = \(\frac{24}{96} = \frac{1}{4}\)
Similar Questions
- An edible oil is sold at the rates 150, 200, 250, 300 rupees per litre in four consecutive years. Assuming that an equal amount of money is spent on oil by a family in every year during these years, what is the average price of oil in rupees (approximately) per litre ?
- Let \(m = 77^n\). The index \(n\) is given a positive integral value at random. What is the probability that the value of \(m\) will have 1 in the units place ?
- What is \(P(A) + P(B) + P(C)\) equal to ?
- A, B, C are three mutually exclusive and exhaustive events associated with a random experiment. If \(3P(B) = 4P(A)\) and \(3P(C) = 2P(B)\), then what is \(P(A)\) equal to?
- A die has two faces with number 4, three faces with number 5 and one face with number 6. If the die is rolled once, then what is the probability of getting 4 or 5?
- A box contains 2 black, 4 yellow and 6 white balls. Three balls are drawn in succession with replacement. What is the probability that all three are of the same colour?
- The letters of the word ZOOLOGY are arranged in all possible ways. What is the probability that the consonants and vowels occur alternatively?
- A natural number \(x\) is chosen at random from the first 100 natural numbers. What is the probability that \(x^2 + x > 50\)?
- A fair coin is tossed till two heads occur in succession. What is the probability that the number of tosses required is less than 6?
- Two perfect dice are thrown. What is the probability that the sum of the numbers on the faces is neither 9 nor 10?
Important Questions from Probability
Two dice are thrown simultaneously. What is the probability of getting the same number on both the dice?
The probability of being 53 Sundays in year 2020 is-
Three dice are thrown randomly. The probability of coming 3 in at least one die is
The probability of having 53 Tuesdays in an ordinary year is:
When two dice are tossed simultaneously, the probability that the sum of the numbers appearing on both the dice is 8 will be
