Question 908034
We're looking for the probability of multiple different scenarios: for example, winning the very first lottery game and no other is one scenario. Winning the second and no other is another scenario. Winning the fourth, fifth, and sixth and no others are yet another scenario. There is a much easier way of solving this problem through trickery, but let's be clear: it will not work for all questions. If this question was asking "What's the probability of winning exactly 1 of 6 games", you would need to calculate the probability of one specific scenario (win game 1), and then multiply by the number of variants (in this case 6).
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If you're taking a test or otherwise in school, this sort of question will generally succumb to simplification -- and that's a reasonably good lesson, as a good first step in solving any question like this, even when it doesn't exist, is to look for a simplifying 'trick' that will reduce the math you have to do. The trick here is that the ONLY situation you're NOT looking for is 'lose every game'. All other outcomes fit the required 'win 1 or more of your 6 plays'.
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So, while the question asks for {{{P(Win Any Games)}}}, you should take it upon yourself to instead look for {{{1 - P(Win No Games)}}}, as that is a much simpler problem.
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From there, it's easy.

{{{P(MultipleEvents) = P(Event1) * P(Event2) * etc.}}}
{{{P(Win No Games) = P(Lose Game 1) * P(Lose Game 2) * etc.}}}
{{{P(Win No Games) = P(Lose Game) ^ (Number Of Games)}}}
{{{P(Win No Games) = (1 - P(Win Game)) ^ 6}}}
{{{P(Win No Games) = (1 - 0.02) ^ 6}}}
{{{P(Win No Games) = 0.98 ^ 6}}}
{{{P(Win No Games) = 0.88584238086}}}
{{{P(Win Any Games) = 1 - 0.98 ^ 6}}}
{{{P(Win Any Games) = 0.11415761913}}}