probability

[jalshri]

Four people each roll a fair die once.Then find the probability that at least two people will roll the same number?
Ans: 13/18

[jgabry]

hey jalshri,

here's one way to approach this question:

Person 1 will roll a die and it will land on some number. So when Person 2 rolls his or her die the probability is 5/6 that the number is NOT the same as person 1's number. Similarly, when Person 3 rolls, the probability will be 4/6 that it lands on neither of the previous two numbers. And likewise, when Person 4 rolls, there will be a 3/6 chance that a unique number shows up.

When multiplied together, these numbers give us the probability that no number comes up twice. If we subtract this from 1, then we will have the probability of the alternative (formally referred to as "the complement") to no number appearing twice, which is that the same number appears 2 or more times.

So, starting with any given number, the probability of three other different numbers = P(different number & different number & different number) = (5/6)(4/6)(3/6) = 5/18

This gives us: P(at least 2 of the same number) = 1 - 5/18 = 18/18 - 5/18 = 13/18.

Hope that helps!

[nareshchowdary28]

Hi Jgabry,

Good Work, Can you suggest me some books or some material where I can really improve my probability skills where I'm lacking after a certain level.
I have Mahattan's word problems book, but as like other books I don't think they have done good job with Probability, probably they should add a topic called difference between Permutations and Combinations and when to use ncr and when to npr.

Please suggest me Jgabry.
Looking forward to get some help.

Naresh

[jgabry]

Hi Naresh,

I'm assuming you mean books/material about the probability and counting (permutations and combinations) concepts that are tested on the GRE, right? If you're interested in learning more about probability in general I definitely make some recommendations for that too, but for now I can suggest a few places to look for GRE-related probability topics:

For permutations and combinations, these two websites are very very basic, but sometimes the easiest examples can provide an insight into the underlying mechanisms that you can then apply to more challenging examples:

http://www.regentsprep.org/Regents/math/algtrig/ATS5/PCPrac.htm

http://betterexplained.com/articles/easy-permutations-and-combinations/

Another resource for gaining an intuition into the differences between permutations and combinations (as well as some probability) is Khan Academy. Here are links some of their videos that you might find helpful:


http://www.youtube.com/watch?v=obZzOq_wSCg&feature=edu&list=PLC58778F28211FA19

http://www.youtube.com/watch?v=QE2uR6Z-NcU&feature=edu&list=PLC58778F28211FA19

http://www.youtube.com/watch?v=mkyZ45KQYi4&feature=edu&list=PLC58778F28211FA19

http://www.youtube.com/watch?v=udG9KhNMKJw&feature=edu&list=PLC58778F28211FA19

http://www.youtube.com/watch?v=XqQTXW7XfYA&feature=edu&list=PLC58778F28211FA19

http://www.youtube.com/watch?v=bCxMhncR7PU&feature=edu&list=PLC58778F28211FA19

http://www.youtube.com/watch?v=Xqfcy1rqMbI&feature=edu&list=PLC58778F28211FA19

http://www.youtube.com/watch?v=W7DmsJKLoxc&feature=edu&list=PLC58778F28211FA19

I hope these are helpful. If you're already beyond the skill required in these videos/articles then you're in very good shape for the GRE. If you're not, then I think these resources can help you get there!

And definitely let me know if you have any other questions.

[tommywallach]

Hey Guys,

Great discussion, and great use of the complement (We often refer to it as "the 1-x trick") to solve the question.

Definitely do not over-prepare on probability or permutations/combinations. The latter almost never shows up on the GRE, and the former is rare. That isn't to say you shouldn't look at it at all, but there are a lot of categories that will pay more dividends in terms of time invested (number properties, algebra, etc.). That being said, it's never bad to be good at everything.

If you're trying to be really advanced, you can always look at GMAT probability/combinatorics questions, which are the same style, but slightly harder. If you can do those, the GRE will be a piece of cake!

-t

[garima_aries01]

Hi
I have a question

If i try doing this question this way
Probability that atleast 2 people get the same number=
probability that exactly 2 people get the same number+
probability that exactly 3 people get the same number+
probability that exactly 4 people get the same number

I should get the same answer , So please make corrections in my approach mentioned :(

probability that exactly 2 people get the same number = 1 x (1/6) x (5/6) x (4/6)

probability that exactly 3 people get the same number = 1 x (1/6) x (1/6) x (5/6)

probability that exactly 4 people get the same number = 1 x (1/6) x (1/6) x (1/6)

adding all i am getting 13/108 :(

I understood the approach mentioned in the above posts but i want to know this approach too please help

[tommywallach]

Hey Garima,

Well, first off, DO NOT DO IT THIS WAY. The 1-x trick is THE ONLY GOOD WAY to do this question. That being said, I'll explain why this method is bad news. Basically, you left out the possibility that you solved only for one combination. I'll explain by looking at your first example:

probability that exactly 2 people get the same number = 1 x (1/6) x (5/6) x (4/6) = 20/216

This is not the possibility of that happening. this is the possibility of, SPECIFICALLY, the first two being the same, and the second two being different. But that is not the only way that two people could get the same number. There are six ways:

1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, and 3 and 4 = 6 ways

So we multiply the probability you found by 6:

20/216 * 6 = 120/216

That's the first possibility. To continue:

probability that exactly 3 people get the same number = 1 x (1/6) x (1/6) x (5/6) = 5/216

123 124 134 234 = 4 ways

5/216 * 4 = 20/216

probability that exactly 4 people get the same number = 1 x (1/6) x (1/6) x (1/6) = 1/216

only way

NOW, another thing you left out is the probability that two sets of two people will get the same things. That would also fit into the category of "at least 2 people get the same thing", so we need to solve for it:

Probability that 2 people get the same number and the other two people get the same number = 1 * 1/6 * 5/6 * 1/6 = 5/216

But just like the others, this can happen in multiple ways:

12 34
13 24
14 23

3 ways -->5/216 * 3 = 15/216

Add all those up and you'll get 156/216 = 13/18

PHEW! See why the 1-x trick is better! : )

-t

[navigalactus]

Tommy is right about the benefits of 1-x method.

At least 2 Persons have to get same number.

So, According to the Adverse situation we are creating, None of the 2 persons have same number which means ALL 3 have Different numbers and 1 of four of them is feeling left out and sad as he couldn't find a match :(

First Person A played his chance by rolling the Dice and got a number. So rest 3 People are left with 5 choices, 4 choices and 3 choices respectively in order to NOT to have a match with Person A's number.

Hence Required Probability =

:)

[tommywallach]

Thanks for the +1, Galactus (and for the awesome name).

-t