A box contains `10` tickets numbered from `1` to `10` . Two tickets are drawn one by one without replacement. The probability that the "difference between the first drawn ticket number and the second is not less than `4" is

Options

A. `(7)/(30)`
B. `(14)/(30)`
C. `(11)/(30)`
D. `(10)/(30)`

Answer

A

Explanation

`(a)` `1234ul(5678910)`
`1^(st)` drawn is `5`, then `2^(nd)` drawn can be `1` only. If `1^(st)` is `6`, then `2^(nd)` is `1` or `2` and so on. ltbr `:. P(E)=(1)/(10)[(1)/(9)+(2)/(9)+(3)/(9)+(4)/(9)+(5)/(9)+(6)/(9)]`
`=(1)/(90)[(6.7)/(2)]`
`=(7)/(30)`