Two boys A and B find the jumble of n ropes lying on the floor. Each takes hold of one loose and randomly. If the probability that they are both holding the same rope is $\frac{1}{101}$ then the number of ropes is equal to

101

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100

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51

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50

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Solution

The correct option is A $101$
A and B have n ropes to choose. So, the number
of ways in which they can choose ropes = $n \times n$
Number of ways in which A and B
choose the same rope = n
Probability = $\frac{n}{n \times n} = \frac{1}{n} = \frac{1}{101}$
$\Rightarrow n = 101$
$∴$ Number of ropes = 101

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Two boys A and B find the jumble of n ropes lying on the floor. Each takes hold of one loose and randomly. If the probability that they are both holding the same rope is 1101 then the number of ropes is equal to

The correct option is A 101A and B have n ropes to choose. So, the numberof ways in which they can choose ropes = n×nNumber of ways in which A and Bchoose the same rope = nProbability = nn×n=1n=1101⇒n=101∴ Number of ropes = 101

Two boys A and B find the jumble of n ropes lying on the floor. Each takes hold of one loose and randomly. If the probability that they are both holding the same rope is $\frac{1}{101}$ then the number of ropes is equal to