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# I know this is not typically a hard one but I have a

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Manager
Joined: 19 Sep 2005
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I know this is not typically a hard one but I have a [#permalink]

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10 Nov 2005, 20:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I know this is not typically a hard one but I have a question about it:

A fair die has sides labeled w/ 1, 2, 3, 4, 5, 6. If the die is rolled 4 times, what is the probability that on at least 1 roll, the die will show 6?

Now, I'd like to know why you HAVE to solve by looking at it from the probability of NOT getting a six. In other words...why can't you add 1/6 4 times? When is it key to go with probability of not getting what you are trying to get, rather than solving for the probability of getting it?

Many thanks
Director
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11 Nov 2005, 00:04
Jennif102 wrote:
I know this is not typically a hard one but I have a question about it:

A fair die has sides labeled w/ 1, 2, 3, 4, 5, 6. If the die is rolled 4 times, what is the probability that on at least 1 roll, the die will show 6?

Now, I'd like to know why you HAVE to solve by looking at it from the probability of NOT getting a six. In other words...why can't you add 1/6 4 times? When is it key to go with probability of not getting what you are trying to get, rather than solving for the probability of getting it?

Many thanks

Jennif102,

the die rolling is a independent event so the total probability will be the multipication of probability of each event.

and the case you are refering where using 1/6 as the probabiltiy of getting 6 .....1/6*1/6*1/6*1/6 .. is the probabiltiy of getting 6 in each event not getting at least one six.

but in the question the probability of at least one 6 is asked .....so in 4 rolls you may have 1 six or 2 six or 3 six ...or all 4 six ......so the best way is to 1 - probability of None= probability of at least one.....otherwise you have to consider the cases all the case

1six3Nonesix+2six2Nonesix...and so on

hope it helps....
Manager
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11 Nov 2005, 07:30
Cool explanation Jonny009.
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11 Nov 2005, 08:12
i will do it as

1/6^4 +2/6^4+ 3/6^4+ 4/6^4

= 1/6^4 ( 1+2+3+4)

= 10/6^4
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hey ya......

Manager
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11 Nov 2005, 17:48
Senior Manager
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11 Nov 2005, 21:10
Here you go:

P(xxxx) + P(6xxx) + P(66xx) + P(666x) + P(6666) = 1

P(6xxx) + P(66xx) + P(666x) + P(6666) = 1 - P(xxxx)

Since each occurance is an independent event.

P(xxxx) = P(x1) * P(x2) * P(x3) * P(x4)

The probability of not getting a 6 is 5/6

P(xxxx) = (5/6)^4

Hence,

P(6xxx) + P(66xx) + P(666x) + P(6666) = 1 - (5/6)^4
11 Nov 2005, 21:10
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