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Dice Proposition (Posted on 2003-02-22) Difficulty: 5 of 5
If a pair of dice are thrown repeatedly until a 7 is thrown, what is the probability that all the other possible numbers from 2 through 12 will have been thrown by the time the 7 has been thrown? As a preliminary, what is the probability that 2, 11 and 12 will all have been thrown before the first 7?

See The Solution Submitted by Charlie    
Rating: 4.2000 (5 votes)

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A try | Comment 4 of 5 |

a) The preliminary problem:

There are 36 possible events; only 1 for numbers 2 and 12, only 2 for number 11; 6 events instead corrispond to number 7.

So only the series with the following forms matches the preliminary problem:

2, 12, 11a, 11b, 7a, 7b, 7c, 7d, 7e, 7f

2, 12, 11a, 7a, ..., 11b, ...7f

2, 12, 11b, 7a, ..., 11a, ...7f

where order could change in the group of terms 2, 12, 11, and also in the group 7a, 7b, 7c, 7d, 7e, 7f.

The first form gives 4!*6! possibilities; the second form gives 3!*6!*6 poss. and the third one also 3!*6!*6. The total space of numbers is constitued by 10! events.

So P= 6!(4+2*3!*6)/10!= 2/105 = 0,01904.

b) The problem posted: the method is the same, but we have to consider all the forms where just one or more events of the same number occurs before the event number 7.

I'll post later.

Edited on May 2, 2005, 6:21 am
  Posted by armando on 2005-05-01 23:14:00

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