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Two natural numbers r, s are drawn one at a time, without replacement from the set \[\mathbf{S}=\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots .,~n \right\}\]. Find \[\mathbf{P}\left[ r~\text{£}~p|s~\text{£}~p \right]\], where \[p\mathbf{\hat{I}}S\].
Two natural numbers r, s are drawn one at a time, without replacement from the set \[\mathbf{S}=\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots .,~n \right\}\]. Find \[\mathbf{P}\left[ r~\text{£}~p|s~\text{£}~p \right]\], where \[p\mathbf{\hat{I}}S\].
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Probability
Two natural numbers r, s are drawn one at a time, without replacement from the set \[\mathbf{S}=\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots .,~n \right\}\]. Find \[\mathbf{P}\left[ r~\text{£}~p|s~\text{£}~p \right]\], where \[p\mathbf{\hat{I}}S\].

Given, \[\mathbf{S}=\left\{ \mathbf{1},\text{ }\mathbf{2},\text{ }\mathbf{3},\text{ }\ldots .,~n \right\}\]

So, \[P\left( r\text{ }\le \text{ }p/s\text{ }\le \text{ }p \right)\text{ }=\text{ }P(P\cap S)/\text{ }P\left( S \right)\]

= \[p\text{ }\text{ }1/n\text{ }\times \text{ }n/\left( n\text{ }\text{ }1 \right)\]

= \[\left( p\text{ }\text{ }1 \right)/\left( n\text{ }\text{ }1 \right)\]

Therefore, the required probability is \[\left( p\text{ }\text{ }1 \right)/\left( n\text{ }\text{ }1 \right)\].

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