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What is the probability of getting no double "4" by rolling

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What is the probability of getting no double "4" by rolling [#permalink] New post 21 Jan 2005, 15:15
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What is the probability of getting no double "4" by rolling two dices?







Which solution is correct:
p(no 4) * p(no 4)
5/6 * 5/6 = 25/36

or

p(no double 4) = 1 - p(double 4)
1 - 1/6 * 1/6 = 35/36

Please help.
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 [#permalink] New post 21 Jan 2005, 15:17
Second is right. The first one means no single 4 is allowed.
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 [#permalink] New post 21 Jan 2005, 16:12
The answer to this will be 5/6 * 5/6 = 25 / 36

that is because they are independent events and in the first roll you want all the numbers execpt 4 so the P of occuring is 5/6

1/6 < 5/6 and the chances of 4 not occuring is more than 4 occuring

hope this helps
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 [#permalink] New post 21 Jan 2005, 16:15
Was I wrong in understanding your question? I thought "no double 4" means you could have a four, as long as you don't get 4 for both rolls.
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 [#permalink] New post 21 Jan 2005, 16:17
no double 4 means no no 4 AND 4 on the 2 die in a single roll
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 [#permalink] New post 21 Jan 2005, 16:18
In other words you could have a 4 and a 5 in a two dice roll, for example. Your answer is still incorrect then.
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 [#permalink] New post 21 Jan 2005, 17:12
Then answer is 35/36.

25/36 is the case when you obtain a 4+X or X+4
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 [#permalink] New post 21 Jan 2005, 23:17
Think about it this way. There are 36 different outcomes for a two dice roll. Only one outcome is excluded (the double four). The probability is 35/36.
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 [#permalink] New post 22 Jan 2005, 06:01
Ops!! I had a 'brain fart' on this one.
35/36 is correct. My bad
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 [#permalink] New post 22 Jan 2005, 09:05
Good question ..............

Twixt could you explain what you mean by 4+x and X +4
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 [#permalink] New post 22 Jan 2005, 09:43
35/36 is correct.

1. 1st four and other not and vice versa = 2 * 5/36 = 10/36

2. No fours atall = 5/6*5/6 = 25/36

Total 35/36
  [#permalink] 22 Jan 2005, 09:43
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