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Maths-

General

Easy

Question

A box contains 100 tickets numbered 1, 2 ...... 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability

$\frac{1}{8}$
$\frac{13}{15}$
$\frac{1}{7}$
None of these

hint Hint:

Formula used
$P left parenthesis B divided by A right parenthesis equals fraction numerator P left parenthesis A intersection B right parenthesis over denominator P left parenthesis A right parenthesis end fraction equals fraction numerator n left parenthesis A intersection B right parenthesis over denominator n left parenthesis A right parenthesis end fraction$

The correct answer is: $fraction numerator 13 over denominator 15 end fraction$

Let $A$ be the event that the maximum number on the two chosen tickets is not more than 10 i.e., the number on them $less or equal than 10$ and $B$ be the event that the maximum number on them is 5, i.e., the number on them is $greater or equal than 5$ we have to find $P left parenthesis B divided by A right parenthesis$ .
Now $P left parenthesis B divided by A right parenthesis equals fraction numerator P left parenthesis A intersection B right parenthesis over denominator P left parenthesis A right parenthesis end fraction equals fraction numerator n left parenthesis A intersection B right parenthesis over denominator n left parenthesis A right parenthesis end fraction$
Now the number of ways of getting a number $r$ on the two tickets is the coefficient of $x to the power of r end exponent$ in the expansion of
$left parenthesis x to the power of 1 end exponent plus x to the power of 2 end exponent plus x to the power of 3 end exponent plus........ plus x to the power of 100 end exponent right parenthesis to the power of 2 end exponent equals x to the power of 2 end exponent left parenthesis 1 plus x plus....... plus x to the power of 99 end exponent right parenthesis to the power of 2 end exponent$
$equals x to the power of 2 end exponent open parentheses fraction numerator 1 minus x to the power of 100 end exponent over denominator 1 minus x end fraction close parentheses to the power of 2 end exponent equals x to the power of 2 end exponent left parenthesis 1 minus 2 x to the power of 100 end exponent plus x to the power of 200 end exponent right parenthesis left parenthesis 1 minus x right parenthesis to the power of negative 2 end exponent$
$equals x to the power of 2 end exponent left parenthesis 1 minus 2 x to the power of 100 end exponent plus x to the power of 200 end exponent right parenthesis left parenthesis 1 plus 2 x plus 3 x to the power of 2 end exponent plus.... plus left parenthesis r plus 1 right parenthesis x to the power of r end exponent plus..... right parenthesis$ Thus coefficient of $x to the power of 2 end exponent equals 1 comma$ of $x to the power of 3 end exponent equals 2 comma$ of $x to the power of 4 end exponent equals 3......$ of $x to the power of 10 end exponent$ is 9.
Hence $n left parenthesis A right parenthesis equals 1 plus 2 plus 3 plus 4 plus 5 plus 6 plus 7 plus 8 plus 9 equals 45$
and $n left parenthesis A intersection B right parenthesis equals 4 plus 5 plus 6 plus 7 plus 8 plus 9 equals 39$
Hence required probability $equals fraction numerator 39 over denominator 45 end fraction equals fraction numerator 13 over denominator 15 end fraction.$

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