- Home
- Standard 11
- Mathematics
14.Probability
hard
Two integers $\mathrm{x}$ and $\mathrm{y}$ are chosen with replacement from the set $\{0,1,2,3, \ldots ., 10\}$. Then the probability that $|x-y|>5$ is:
A
$\frac{30}{121}$
B
$\frac{62}{121}$
C
$\frac{60}{121}$
D
$\frac{31}{121}$
(JEE MAIN-2024)
Solution
$ \text { If } x=0, y=6,7,8,9,10 $
$ \text { If } x=1, y=7,8,9,10 $
$ \text { If } x=2, y=8,9,10 $
$ \text { If } x=3, y=9,10 $
$ \text { 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 $=\frac{30}{11 \times 11}=\frac{30}{121}$
Standard 11
Mathematics
