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If two different numbers are taken from the set {0, 1, 2, 3, ......, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is
- a)7/55
- b)6/55
- c)12/55
- d)14/45
Correct answer is option 'B'. Can you explain this answer?
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If two different numbers are taken from the set {0, 1, 2, 3, ......, 1...
Solution:
Given set of numbers is {0, 1, 2, 3, ......, 10}
Let the two numbers be a and b. Then, we have to find the probability that (a + b) and |a - b| are both multiples of 4.
Case 1: When both a and b are even numbers
There are 6 even numbers in the given set of 11 numbers. So, the probability of selecting an even number as the first number is 6/11. Once the first even number is selected, there are 5 even numbers left out of 10 numbers. So, the probability of selecting the second even number is 5/10. Therefore, the probability of selecting two even numbers is (6/11) x (5/10) = 3/11.
Now, the sum of two even numbers is always an even number and the absolute difference of two even numbers is always a multiple of 4. So, both conditions are satisfied.
Case 2: When both a and b are odd numbers
There are 5 odd numbers in the given set of 11 numbers. So, the probability of selecting an odd number as the first number is 5/11. Once the first odd number is selected, there are 4 odd numbers left out of 10 numbers. So, the probability of selecting the second odd number is 4/10. Therefore, the probability of selecting two odd numbers is (5/11) x (4/10) = 2/11.
Now, the sum of two odd numbers is always an even number and the absolute difference of two odd numbers is always an even number but not necessarily a multiple of 4. So, this case does not satisfy the condition.
Case 3: When one number is even and the other is odd
There are 6 even numbers and 5 odd numbers in the given set of 11 numbers. So, the probability of selecting an even number as the first number is 6/11 and the probability of selecting an odd number as the second number is 5/10. Therefore, the probability of selecting one even and one odd number is (6/11) x (5/10) = 3/11.
Now, the sum of an even number and an odd number is always an odd number and the absolute difference of an even number and an odd number is always an odd number but not necessarily a multiple of 4. So, this case does not satisfy the condition.
Therefore, the only case that satisfies both conditions is case 1.
Hence, the required probability is 3/11.
Therefore, option B is correct.
Given set of numbers is {0, 1, 2, 3, ......, 10}
Let the two numbers be a and b. Then, we have to find the probability that (a + b) and |a - b| are both multiples of 4.
Case 1: When both a and b are even numbers
There are 6 even numbers in the given set of 11 numbers. So, the probability of selecting an even number as the first number is 6/11. Once the first even number is selected, there are 5 even numbers left out of 10 numbers. So, the probability of selecting the second even number is 5/10. Therefore, the probability of selecting two even numbers is (6/11) x (5/10) = 3/11.
Now, the sum of two even numbers is always an even number and the absolute difference of two even numbers is always a multiple of 4. So, both conditions are satisfied.
Case 2: When both a and b are odd numbers
There are 5 odd numbers in the given set of 11 numbers. So, the probability of selecting an odd number as the first number is 5/11. Once the first odd number is selected, there are 4 odd numbers left out of 10 numbers. So, the probability of selecting the second odd number is 4/10. Therefore, the probability of selecting two odd numbers is (5/11) x (4/10) = 2/11.
Now, the sum of two odd numbers is always an even number and the absolute difference of two odd numbers is always an even number but not necessarily a multiple of 4. So, this case does not satisfy the condition.
Case 3: When one number is even and the other is odd
There are 6 even numbers and 5 odd numbers in the given set of 11 numbers. So, the probability of selecting an even number as the first number is 6/11 and the probability of selecting an odd number as the second number is 5/10. Therefore, the probability of selecting one even and one odd number is (6/11) x (5/10) = 3/11.
Now, the sum of an even number and an odd number is always an odd number and the absolute difference of an even number and an odd number is always an odd number but not necessarily a multiple of 4. So, this case does not satisfy the condition.
Therefore, the only case that satisfies both conditions is case 1.
Hence, the required probability is 3/11.
Therefore, option B is correct.
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If two different numbers are taken from the set {0, 1, 2, 3, ......, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, isa)7/55b)6/55c)12/55d)14/45Correct answer is option 'B'. Can you explain this answer?
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