A) \[\frac{81}{121}\]
B) \[\frac{30}{121}\]
C) \[\frac{25}{121}\]
D) \[\frac{20}{121}\]
Correct Answer: B
Solution :
[b]: The total number of selections of two numbers x and y from the numbers 0 to 10 \[=11\times 11=121\] Now, if \[|x-y|>5\], then the possible values for (x, y) are (0, 6), (0, 7), (0, 8), (0, 9), (0 10), (1, 7), (1, 8), (1, 9), (1, 10), (2, 8), (2, 9), (2, 10), (3, 9'), (3, 10), (4, 10), (6, 0), (7, 0), (8, 0), (9, 0), (10, 0), (7, 1), (8, 1), (9, 1), (8, 2), (10, 1), (9, 2), (10, 2), (9, 3), (10,3), (10, 4) So, there are 30 pairs of values of x and) y. Hence, the required probability\[=\frac{30}{121}\].
You need to login to perform this action.
You will be redirected in
3 sec
