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Simple Probability Problem. Help with concept needed :-)

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Simple Probability Problem. Help with concept needed :-)

by amazonviper » Sat May 16, 2009 6:05 pm
Problem:
A dice is rolled 6 times. What is the Probability that "6" will NOT be facing upward on all three rolls ?.

My method:

Prob that we will not get a "6" first time = 5/6
Prob that we will not get a "6" 2nd time = 5/6
Prob that we will not get a "6" 3rd time = 5/6

So, Prob that we will not get "6" all three times is 5/6*5/6*5/6

Seems that this is incorrect.The book calculates Probability of 6 appearing all three times and then subtracting from 1 to get the answer.Why cannot we use the method above(My method)?

To me, this type of question seems straightforward and I want to be sure that I am not using the wrong logic here

Thanks in advance for the help :-)

Viper
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Hello!

by amazonviper » Mon May 18, 2009 1:44 pm
Any help is greatly appreciated!
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by ManSab » Mon May 18, 2009 3:34 pm
total outcomes in 3 rolls = 18
6 can face up 3 times in 3 rolls.

probability for 6 facing up in 3 rolls = 3/18 = 1/6
probability 6 will not face = 1- 1/6 = 5/6.

Please correct if I am wrong.

Thanks
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by dumb.doofus » Tue May 19, 2009 3:57 pm
ok.. well here's the problem with your solution.. and first of all thanks for asking this question.. quite a conceptual topic..

when you say that the probability of not getting a 6 is 5/6, the first time and second time and third time.. essentially the you are finding all combinations where 6 is not used at all.. examples are:

1 2 3
3 4 5
4 5 2
1 3 5
5 3 1 and I could go on and on 125 times.. that is what your answer is...

But this answer doesn't take into account the following choices:

1 3 6
6 4 5
6 4 2
3 6 5 and so on.. 90 of them to be precise.

Now the reason why your solution is not taking into account these choices.. the problem is that you have already mentioned that there are 5 ways to choose for first number, 2nd number and 3rd number.. and that's the reason you are getting 125 ways.. we also have to exclude the above 90 ways tooo

So the correct solution is:

we know that there is only one combination out of 216 combinations, where we will get 6 on all the three rolls.. i.e. 6 6 6

so the probability of getting a 6 on all three is 1/216

so the probability of not getting a 6 at all is 1 - 1/216 = 215/216


Hope this clarifies..
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by mike22629 » Wed May 20, 2009 11:26 am
The 1-x probability is the best way to answer this problem.

Learn how to do this because it will make probability problems on the GMAT a lot easier.

First, find total possibilities of rolls.

6^3 = 216

First roll = 6 possibilities
Second roll = 6 possibilites
Third Roll = 6 possibilites

Now find the total number of possibilities of rolling 6 all three times.

Obviously the answer is 1

So....
216-1 = 215 (possibilities of not rolling 6 all three times)

Answer 215/216

Trust me when I tell you that this is VERY COMMON in probability problems on GMAT. It is worth your time to master the "1-x" probability approach.
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