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
A
`(7)/(30)`
B
`(14)/(30)`
C
`(11)/(30)`
D
`(10)/(30)`
Text Solution
Verified by Experts
The correct Answer is:
A
`(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)`
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