Is it possible to assign a "run probability" to any function?

fn = RandomChoice[{#3, 1 - #3} -> {RandomSample[#1, #2], {}}] &;

Use:

fn[list,number_of_samples,probability_of_sample_success]

Example:

fn[{1, 2, 3, 4, 5, 6}, 3, .3]

{2,1,5}

fn[{1, 2, 3, 4, 5, 6}, 3, .3]

{1,4,5}

fn[{1, 2, 3, 4, 5, 6}, 3, .3]

{}

You can use RandomChoice to do weighted random choice of the sampling functions and use Through to apply those functions to your data.

d = {1, 2, 3, 4, 5, 6};

SeedRandom[15]
Through[RandomChoice[{0.7, 0.3} -> {RandomSample[#, 1] &, Nothing &}, 12][d]]

(* {{3}, {5}, {1}, {6}, {1}, {4}, {4}, {3}, {3}} *)

SeedRandom[15]
Through[RandomChoice[{0.7, 0.3} -> {RandomSample[#, 1] &, None &}, 12][d]]

(* {None, None, {3}, {5}, {1}, {6}, {1}, {4}, {4}, None, {3}, {3}} *)

Perhaps this:

d = {1, 2, 3, 4, 5, 6};

RandomChoice[
 Flatten@{70, ConstantArray[30/Length[d], Length[d]]} ->
  {{}, Sequence @@ d}
]

Let's check that it does what you want by drawing 10,000 samples:

RandomChoice[
  Flatten@{70, ConstantArray[30/Length[d], Length[d]]} ->
   {{}, Sequence @@ d},
  10000
] // Counts

(* Out: <|{} -> 6998, 3 -> 478, 2 -> 481, 1 -> 489, 5 -> 494, 4 -> 554, 6 -> 506|> *) 

The frequency of {} (i.e. the "no sample" response) is 6998/10000, i.e.pretty close to 70%. The others are pretty evenly distributed as well.