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A fair die is rolled 12 times. Consider the following four possible outcomes: (i) 5 2 6 3 2 1 4 1 6 5 3 4 (ii) 1 1 2 2 3 3 4 4 5 5 6 6 (iii) 6 6 6 6 6 6 6 6 6 6 6 6 (iv) 1 5 4 3 5 1 2 4 4 6 4 5 Which of the following is the most likely outcome: (i), (ii), (iii), (iv)? (i) because number of die outcomes (1, 2, 3, 4, 5, 6) is equal but in a random order. (ii) because the number of die outcomes (1, 2, 3, 4, 5, 6) is equal. (iii) because the number 6 is just as likely as any other number on a die. (iv) because you won’t necessarily get the same number of of die outcomes (1, 2, 3, 4, 5, 6) with a fair die. They are all equally likely.

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Answer: Hello mate!

A fair die has the possible results of 1, 2, 3, 4, 5 and 6, where all have the same probability ( that is 1/6).

then when you throw the dice 12 times, all the possible arrays of 12 numbers between 1 and 6 have the same probability of showing up, that is equal to

(1/6)^12

this means that the distribution (iii) 6 6 6 6 6 6 6 6 6 6 6 6 (where each 6 has a 1/6 probability) has the same probabilities that  (iv) 1 5 4 3 5 1 2 4 4 6 4 5 (where each number has the same probability; 1/6) and we here are imposing order, so the first number must be a 1, which has a 1/6, the second must be a 5, which also has a 1/6; this is the same for every number (doesn't matter if they are different or equal)

then the correct answer is: with a fair die. They are all equally likely.

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