Question

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.

Answer 1

anyone? i need a walk through on how to do this... please

Answer 2

i know each number has a probability of 1/6

Answer 3

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.

Answer 4

oh my bad.... The probability that each player has of winning is .....?

Answer 5

How exactly do people win in this game show?

Answer 6

its kinda like the lottery i guess... you predict numbers and if your numbers are called then, you win

Answer 7

sorry you receive numbers

Answer 8

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.

Answer 9

*I meant 1/36

Answer 10

well the choices are:

1/2

1/3

2/3

1/12

Answer 11

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.

Answer 12

seems probable... haha ill go for it, thanks

Answer 13

it was 1/12 haha @Cardinal_Carlo