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  • Two integers are chosen (without random) at random from the s

Two integers \mathrm{x \: and \: y} are chosen (without random) at random from the set \mathrm{\{x: 0 \leq x \leq 10, x} is an integer}. If the probability for \mathrm{|x-y| \leq 5\: is\: p}, then the value of \mathrm{121 p} must be

Option: 1

91


Option: 2

90


Option: 3

1


Option: 4

121


Answers (1)

best_answer

There are 11 ways to choose $x$ and 11 ways to choose \mathrm{y}. If \mathrm{S} be the sample space, then

\mathrm{n(S) =} Total number of choosing  \times and \mathrm{ y }
\mathrm{ =11 \times 11=121}

The number of different values of \mathrm{y} for a given value of \mathrm{x} can be determined as follows
when \mathrm{x=0}, we have \mathrm{|0-y| \leq 5}

\mathrm{\Rightarrow |y| \leq 5 }

\mathrm{\Rightarrow -5 \leq y \leq 5 }

\mathrm{ \Rightarrow 0 \leq y \leq 5 }

Gives six values of \mathrm{ y, ie, \{0,1,2,3,4,5\} }

\mathrm{ \{\because y \geq 0}

When \mathrm{ x=1}, we have \mathrm{ |1-y| \leq 5}

\mathrm{ \Rightarrow -5 \leq 1-y \leq 5 }

\mathrm{ \Rightarrow 5 \geq y-1 \geq-5 }

\mathrm{ \Rightarrow 6 \geq y \geq-4 }

\mathrm{ \Rightarrow 0 \leq y \leq 6 }

\mathrm{ \Rightarrow 0 \leq y \leq 6 }                            {because \mathrm{ y\geq 0}}

Gives seven values of \mathrm{ y, i e,\{0,1,2,3,4,5,6\} When\: \: x=2}, we have

\mathrm{ |2-y| \leq 5 }

\mathrm{ \Rightarrow -5 \leq 2-y \leq 5 }

\mathrm{ \Rightarrow 5 \geq-2+y \geq-5}

\mathrm{ \Rightarrow 7 \geq y \geq-3 \\ }

\mathrm{ \Rightarrow 0 \leq y \leq 7 }
(Since \mathrm{ y \geq 0 } )
Gives 8 values of \mathrm{ y, ie, \{0,1,2,3,4,5,6,7\} } similarly we can show that when \mathrm{x} equals 3,4,5,6,7,8,9,10 there are 9,10,11,10,9,8,7,6 ; \mathrm{y}-values respectively. Let \mathrm{E} be the event of favourable cases, then

\mathrm{ n(E) =6+7+8+9+10+11+10+9+8+7+6 }

\mathrm{ =91 }

Hence, required probability,

\mathrm{ P(E)=\frac{n(E)}{n(S)} =\frac{91}{121}=p \text { (given) } }

\mathrm{ \therefore 121 p =91 }

Hence option 1 is correct.







 

Posted by

Gaurav

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