let x and y be random numbers from 0 to 10 where x,y∈R ∣x−y∣≥d what is the probability that their sum is less than 10 in the following cases a) d=0 b) d=1 c) d=2


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Question Number 205394 by Red1ight last updated on 19/Mar/24

$let x and y be random numbers from 0 to 10$ $where x, y \in R$ $∣ x - y ∣⩾ d$ $what is the probability that their sum is less than 10$ $in the following cases$ $a) d = 0$ $b) d = 1$ $c) d = 2$
	Answered by mr W last updated on 20/Mar/24

	Commented by mr W last updated on 20/Mar/24
	$∣x−y∣≥d ⇒ { ((x−y≥d ⇒y≤x−d)),((or)),((x−y≤−d ⇒y≥x+d)) :} ∣x−y∣≤d means when point lies in hatched zones. x+y≤10 ⇒y≤10−x ⇒shaded area p=(((10−d)^2 )/(2×10^2 ))=(1/2)(1−(d/(10)))^2 p_(d=0) =0.5 p_(d=1) =0.405 p_(d=2) =0.32$
	$∣ x - y ∣⩾ d$ $\Rightarrow {\begin{cases} x - y ⩾ d \Rightarrow y ⩽ x - d \\ o r \\ x - y ⩽ - d \Rightarrow y ⩾ x + d \end{cases}$ $∣ x - y ∣⩽ d m e a n s w h e n p o i n t l i e s i n$ $h a t c h e d z o n e s .$ $x + y ⩽ 10 \Rightarrow y ⩽ 10 - x \Rightarrow s h a d e d a r e a$ $p = \frac{{(10 - d)}^{2}}{2 \times 10^{2}} = \frac{1}{2} {(1 - \frac{d}{10})}^{2}$ $p_{d = 0} = 0.5$ $p_{d = 1} = 0.405$ $p_{d = 2} = 0.32$
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