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Two-integers-x-and-y-are-chosen-with-replacement-out-of-the-set-0-1-2-3-10-Then-the-probability-that-x-y-gt-5-is-




Question Number 26851 by jain50791@gmail.com last updated on 30/Dec/17
Two integers x and y are chosen with  replacement out of the set {0, 1, 2, 3,..., 10}.  Then the probability that ∣x−y∣>5 is
$$\mathrm{Two}\:\mathrm{integers}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{with} \\ $$$$\mathrm{replacement}\:\mathrm{out}\:\mathrm{of}\:\mathrm{the}\:\mathrm{set}\:\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},…,\:\mathrm{10}\right\}. \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mid{x}−{y}\mid>\mathrm{5}\:\mathrm{is} \\ $$
Commented by Rasheed.Sindhi last updated on 30/Dec/17
Let x>y  ∣x−y∣>5⇒x−y>5⇒x>y+5  y=0⇒x=6,7,...,10 (5 values)       [5  Satisfying pairs]  y=1⇒x=7,8,...,10  (4 values)  y=2⇒x=8,...,10  (3 values)  y=3⇒x=9,10  (2 values)  y=4⇒x=10  (1 values)  Total satifying pairs: 15 for x>y  ∵ (a,b) satisfy⇒(b,a) also satisfy  Total satifying pairs: 15 for x<y  [Note that for x=y no pair satisfy]  Hence number of all pairs which  satisfy ∣x−y∣>5 is 15+15=30  Number of all possible pairs of the  domain set     =11×11=121  Probability=((Number of satisfying pairs)/(Number of total pairs))  Probability=((30)/(121))
$$\mathrm{Let}\:\mathrm{x}>\mathrm{y} \\ $$$$\mid\mathrm{x}−\mathrm{y}\mid>\mathrm{5}\Rightarrow\mathrm{x}−\mathrm{y}>\mathrm{5}\Rightarrow\mathrm{x}>\mathrm{y}+\mathrm{5} \\ $$$$\mathrm{y}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{6},\mathrm{7},…,\mathrm{10}\:\left(\mathrm{5}\:\mathrm{values}\right) \\ $$$$\:\:\:\:\:\left[\mathrm{5}\:\:\mathrm{Satisfying}\:\mathrm{pairs}\right] \\ $$$$\mathrm{y}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{7},\mathrm{8},…,\mathrm{10}\:\:\left(\mathrm{4}\:\mathrm{values}\right) \\ $$$$\mathrm{y}=\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{8},…,\mathrm{10}\:\:\left(\mathrm{3}\:\mathrm{values}\right) \\ $$$$\mathrm{y}=\mathrm{3}\Rightarrow\mathrm{x}=\mathrm{9},\mathrm{10}\:\:\left(\mathrm{2}\:\mathrm{values}\right) \\ $$$$\mathrm{y}=\mathrm{4}\Rightarrow\mathrm{x}=\mathrm{10}\:\:\left(\mathrm{1}\:\mathrm{values}\right) \\ $$$$\mathrm{Total}\:\mathrm{satifying}\:\mathrm{pairs}:\:\mathrm{15}\:\mathrm{for}\:\mathrm{x}>\mathrm{y} \\ $$$$\because\:\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{satisfy}\Rightarrow\left(\mathrm{b},\mathrm{a}\right)\:\mathrm{also}\:\mathrm{satisfy} \\ $$$$\mathrm{Total}\:\mathrm{satifying}\:\mathrm{pairs}:\:\mathrm{15}\:\mathrm{for}\:\mathrm{x}<\mathrm{y} \\ $$$$\left[\mathrm{Note}\:\mathrm{that}\:\mathrm{for}\:\mathrm{x}=\mathrm{y}\:\mathrm{no}\:\mathrm{pair}\:\mathrm{satisfy}\right] \\ $$$$\mathrm{Hence}\:\mathrm{number}\:\mathrm{of}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{which} \\ $$$$\mathrm{satisfy}\:\mid\mathrm{x}−\mathrm{y}\mid>\mathrm{5}\:\mathrm{is}\:\mathrm{15}+\mathrm{15}=\mathrm{30} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{domain}\:\mathrm{set} \\ $$$$\:\:\:=\mathrm{11}×\mathrm{11}=\mathrm{121} \\ $$$$\mathrm{Probability}=\frac{\mathrm{Number}\:\mathrm{of}\:\mathrm{satisfying}\:\mathrm{pairs}}{\mathrm{Number}\:\mathrm{of}\:\mathrm{total}\:\mathrm{pairs}} \\ $$$$\mathrm{Probability}=\frac{\mathrm{30}}{\mathrm{121}} \\ $$

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