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Two numbers a and b are chosen at random from the set of first 30 natural numbers. The probability that a2 - b2 is divisible by 3 is
- a)9/87
- b)12/87
- c)38/87
- d)47/87
Correct answer is option 'D'. Can you explain this answer?
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Two numbers a and b are chosen at random from the set of first 30 natu...
To find the probability that a² - b² is divisible by 3, we need to consider the different cases where a and b can be chosen.
Case 1: Both a and b are divisible by 3
In this case, there are 10 numbers (3, 6, 9, ..., 30) that are divisible by 3. The number of ways to choose 2 numbers from this set is given by the combination formula:
C(10, 2) = 10! / (2! * (10-2)!) = 45
Case 2: Both a and b leave a remainder of 1 when divided by 3
In this case, there are 10 numbers (1, 4, 7, ..., 28) that leave a remainder of 1 when divided by 3. The number of ways to choose 2 numbers from this set is also given by the combination formula:
C(10, 2) = 10! / (2! * (10-2)!) = 45
Case 3: Both a and b leave a remainder of 2 when divided by 3
In this case, there are 10 numbers (2, 5, 8, ..., 29) that leave a remainder of 2 when divided by 3. The number of ways to choose 2 numbers from this set is again given by the combination formula:
C(10, 2) = 10! / (2! * (10-2)!) = 45
Case 4: One number is divisible by 3 and the other leaves a remainder of 1
In this case, there are 10 numbers that are divisible by 3 and 10 numbers that leave a remainder of 1 when divided by 3. The number of ways to choose 1 number from each set is given by the multiplication principle:
10 * 10 = 100
Case 5: One number is divisible by 3 and the other leaves a remainder of 2
In this case, there are 10 numbers that are divisible by 3 and 10 numbers that leave a remainder of 2 when divided by 3. The number of ways to choose 1 number from each set is again given by the multiplication principle:
10 * 10 = 100
Case 6: One number leaves a remainder of 1 and the other leaves a remainder of 2
In this case, there are 10 numbers that leave a remainder of 1 when divided by 3 and 10 numbers that leave a remainder of 2 when divided by 3. The number of ways to choose 1 number from each set is given by the multiplication principle:
10 * 10 = 100
Total number of ways to choose two numbers from the set of first 30 natural numbers = C(30, 2) = 30! / (2! * (30-2)!) = 435
Therefore, the probability that a² - b² is divisible by 3 is given by the sum of the probabilities of each case divided by the total number of ways:
P = (45 + 45 + 45 + 100 + 100 + 100) / 435 = 47/87
Hence, the correct answer is option D) 47/87.
Case 1: Both a and b are divisible by 3
In this case, there are 10 numbers (3, 6, 9, ..., 30) that are divisible by 3. The number of ways to choose 2 numbers from this set is given by the combination formula:
C(10, 2) = 10! / (2! * (10-2)!) = 45
Case 2: Both a and b leave a remainder of 1 when divided by 3
In this case, there are 10 numbers (1, 4, 7, ..., 28) that leave a remainder of 1 when divided by 3. The number of ways to choose 2 numbers from this set is also given by the combination formula:
C(10, 2) = 10! / (2! * (10-2)!) = 45
Case 3: Both a and b leave a remainder of 2 when divided by 3
In this case, there are 10 numbers (2, 5, 8, ..., 29) that leave a remainder of 2 when divided by 3. The number of ways to choose 2 numbers from this set is again given by the combination formula:
C(10, 2) = 10! / (2! * (10-2)!) = 45
Case 4: One number is divisible by 3 and the other leaves a remainder of 1
In this case, there are 10 numbers that are divisible by 3 and 10 numbers that leave a remainder of 1 when divided by 3. The number of ways to choose 1 number from each set is given by the multiplication principle:
10 * 10 = 100
Case 5: One number is divisible by 3 and the other leaves a remainder of 2
In this case, there are 10 numbers that are divisible by 3 and 10 numbers that leave a remainder of 2 when divided by 3. The number of ways to choose 1 number from each set is again given by the multiplication principle:
10 * 10 = 100
Case 6: One number leaves a remainder of 1 and the other leaves a remainder of 2
In this case, there are 10 numbers that leave a remainder of 1 when divided by 3 and 10 numbers that leave a remainder of 2 when divided by 3. The number of ways to choose 1 number from each set is given by the multiplication principle:
10 * 10 = 100
Total number of ways to choose two numbers from the set of first 30 natural numbers = C(30, 2) = 30! / (2! * (30-2)!) = 435
Therefore, the probability that a² - b² is divisible by 3 is given by the sum of the probabilities of each case divided by the total number of ways:
P = (45 + 45 + 45 + 100 + 100 + 100) / 435 = 47/87
Hence, the correct answer is option D) 47/87.
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Two numbers a and b are chosen at random from the set of first 30 natural numbers. The probability that a2 - b2 is divisible by 3 isa)9/87b)12/87c)38/87d)47/87Correct answer is option 'D'. Can you explain this answer?
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