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December 20, 2014

December 20, 2014

Posted by **Ed** on Saturday, May 7, 2011 at 3:39pm.

The answer is 3/11. Could someone tell me how to get this. Thanks?

- Math -
**PsyDAG**, Sunday, May 8, 2011 at 2:57pmThe rolls are independent. If you have already obtained a 1, you are asking for the probability of getting either a 1 or 2 to add to "less than 4."

1/6 + 1/6 = 2/6 = 1/3

With 6 possibilities for any roll of a die, there is no way of getting a denominator of 11.

http://www.members.cox.net/dagershaw/lol/Odds.html

- Math -
**PsyDAGisDumb**, Thursday, June 19, 2014 at 10:55pmPossibilities with rolling 3 on ONE of the rolls:

First roll: 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6

Second roll: 1,1 | 2,1 | 3,1 | 4,1 | 5,1 | 6,1

Eliminate 1,1 from second roll because it is the same as in the first, therefore there are 11 possibilities with 1 as either the first or second roll.

Find the rolls that are less than 4.

(1,1) (1,2) (2,1). 3 rolls are less than 4.

Therefore P(x < 4) = 3 / 11.

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