anonymous
  • anonymous
Contestants in a game show are each given three pairs of numbers (for example, 2 and 3, 5 and 6, and 6 and 1). A person in the audience is randomly picked and asked to roll two dice, one white and one red. The white die gives the first number in the pair, and the red die gives the second number.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
anyone? i need a walk through on how to do this... please
anonymous
  • anonymous
i know each number has a probability of 1/6
anonymous
  • anonymous
A dice has 6 sides or 6 possible values (1-6). You were given two sets: the first set {2, 5 , 6} and a second {3, 6, 1}. Although, yes, you are right that each number has a probability of 1/6, an actual question isn't posted above so we can't really help you.

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anonymous
  • anonymous
oh my bad.... The probability that each player has of winning is .....?
anonymous
  • anonymous
How exactly do people win in this game show?
anonymous
  • anonymous
its kinda like the lottery i guess... you predict numbers and if your numbers are called then, you win
anonymous
  • anonymous
sorry you receive numbers
anonymous
  • anonymous
Never mind that. If we assume that getting chosen by both the white and red dice is how a player wins, then we can calculate the odds of that ever happening. As you said earlier, each numeric value has a 1/6 probability of occurring. Since two simultaneous events (i.e. two dices) determine the odds of winning, we can square our original probability to get an actual probability. \[\left( \frac{ 1 }{ 6 } \right)^{2} = \frac{ 1 }{ 36 }\] In simpler terms, if the winning criteria is having both numerical pairs selected, then each player's odds are ubiquitously 1/32.
anonymous
  • anonymous
*I meant 1/36
anonymous
  • anonymous
well the choices are: 1/2 1/3 2/3 1/12
anonymous
  • anonymous
Hmm, so the odds of winning are not determined by both dices at the same time, but rather, it is determine by both dices at separate probabilities. In that case, we double the probability instead of squaring it. \[\frac{ 1 }{ 6 } + \frac{ 1 }{ 6 } = \frac{ 2 }{ 6 } = \frac{ 1 }{ 3 }\] In other words, each player has 1/3 probability of being selected by either dice. And not by both simultaneously, as I had mistakenly assumed earlier.
anonymous
  • anonymous
seems probable... haha ill go for it, thanks
anonymous
  • anonymous
it was 1/12 haha @Cardinal_Carlo

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