JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 13)
Two integers $$x$$ and $$y$$ are chosen with replacement from the set $$\{0,1,2,3, \ldots, 10\}$$. Then the probability that $$|x-y|>5$$, is :
$$\frac{31}{121}$$
$$\frac{60}{121}$$
$$\frac{62}{121}$$
$$\frac{30}{121}$$
Explanation
If $$x=0, y=6,7,8,9,10$$
If $$x=1, y=7,8,9,10$$
If $$x=2, y=8,9,10$$
If $$x=3, y=9,10$$
If $$x=4, y=10$$
If $$x=5, y=$$ no possible value
Total possible ways $$=(5+4+3+2+1) \times 2$$
$$=30$$
Required probability $$=\frac{30}{11 \times 11}=\frac{30}{121}$$
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